Implicit Differentiation Bernd Schr oder logo1 Bernd Schr oder - - PowerPoint PPT Presentation

logo1 Introduction Derivatives Related Rates Inverse Functions

Implicit Differentiation

Bernd Schr¨

• der

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

(And some equations cannot be solved symbolically.)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

(And some equations cannot be solved symbolically.)

• 3. But note that y = y(x) is a function in the equation above.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

(And some equations cannot be solved symbolically.)

• 3. But note that y = y(x) is a function in the equation above.
• 4. Because equal functions have equal derivatives, we can

take the derivative of both sides of the equation and get a valid new equation.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

(And some equations cannot be solved symbolically.)

• 3. But note that y = y(x) is a function in the equation above.
• 4. Because equal functions have equal derivatives, we can

take the derivative of both sides of the equation and get a valid new equation.

• 5. The main challenge is to remember that y is a function

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

(And some equations cannot be solved symbolically.)

• 3. But note that y = y(x) is a function in the equation above.
• 4. Because equal functions have equal derivatives, we can

take the derivative of both sides of the equation and get a valid new equation.

• 5. The main challenge is to remember that y is a function,

which means that we must use the chain rule to differentiate terms with y.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

(And some equations cannot be solved symbolically.)

• 3. But note that y = y(x) is a function in the equation above.
• 4. Because equal functions have equal derivatives, we can

take the derivative of both sides of the equation and get a valid new equation.

• 5. The main challenge is to remember that y is a function,

which means that we must use the chain rule to differentiate terms with y. That is, d dxh(y)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

(And some equations cannot be solved symbolically.)

• 3. But note that y = y(x) is a function in the equation above.
• 4. Because equal functions have equal derivatives, we can

take the derivative of both sides of the equation and get a valid new equation.

• 5. The main challenge is to remember that y is a function,

which means that we must use the chain rule to differentiate terms with y. That is, d dxh(y) = d dxh

• y(x)
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

(And some equations cannot be solved symbolically.)

• 3. But note that y = y(x) is a function in the equation above.
• 4. Because equal functions have equal derivatives, we can

take the derivative of both sides of the equation and get a valid new equation.

• 5. The main challenge is to remember that y is a function,

which means that we must use the chain rule to differentiate terms with y. That is, d dxh(y) = d dxh

• y(x)
• = h′

y(x) d dxy(x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

(And some equations cannot be solved symbolically.)

• 3. But note that y = y(x) is a function in the equation above.
• 4. Because equal functions have equal derivatives, we can

take the derivative of both sides of the equation and get a valid new equation.

• 5. The main challenge is to remember that y is a function,

which means that we must use the chain rule to differentiate terms with y. That is, d dxh(y) = d dxh

• y(x)
• = h′

y(x) d dxy(x) = h′ y(x)

• y′(x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Derivatives of Inconvenient Functions

• 1. What would we do if we had to find the derivative of the

function y that satisfies y3 +3y2 −4 = x2?

• 2. It is possible to solve for y. The formula is quite horrible.

(And some equations cannot be solved symbolically.)

• 3. But note that y = y(x) is a function in the equation above.
• 4. Because equal functions have equal derivatives, we can

take the derivative of both sides of the equation and get a valid new equation.

• 5. The main challenge is to remember that y is a function,

which means that we must use the chain rule to differentiate terms with y. That is, d dxh(y) = d dxh

• y(x)
• = h′

y(x) d dxy(x) = h′ y(x)

• y′(x) = h′(y)y′.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2y′(x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2y′(x)+6

• y(x)
• y′(x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2y′(x)+6

• y(x)
• y′(x)

= 2x

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2y′(x)+6

• y(x)
• y′(x)

= 2x 3y2y′ +6yy′ = 2x

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2y′(x)+6

• y(x)
• y′(x)

= 2x 3y2y′ +6yy′ = 2x y′ 3y2 +6y

• =

2x

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2y′(x)+6

• y(x)
• y′(x)

= 2x 3y2y′ +6yy′ = 2x y′ 3y2 +6y

• =

2x y′ = 2x 3y2 +6y

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2y′(x)+6

• y(x)
• y′(x)

= 2x 3y2y′ +6yy′ = 2x y′ 3y2 +6y

• =

2x y′ = 2x 3y2 +6y y′

• (0,1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2y′(x)+6

• y(x)
• y′(x)

= 2x 3y2y′ +6yy′ = 2x y′ 3y2 +6y

• =

2x y′ = 2x 3y2 +6y y′

• (0,1)

= 2x 3y2 +6y

• (0,1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2y′(x)+6

• y(x)
• y′(x)

= 2x 3y2y′ +6yy′ = 2x y′ 3y2 +6y

• =

2x y′ = 2x 3y2 +6y y′

• (0,1)

= 2x 3y2 +6y

• (0,1)

= 0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. The function y satisfies y3 +3y2 −4 = x2. Find its

derivative at the point (0,1).

• y(x)

3 +3

• y(x)

2 −4 = x2 d dx

• y(x)

3 +3

• y(x)

2 −4

• =

d dx

• x2

3

• y(x)

2y′(x)+6

• y(x)
• y′(x)

= 2x 3y2y′ +6yy′ = 2x y′ 3y2 +6y

• =

2x y′ = 2x 3y2 +6y y′

• (0,1)

= 2x 3y2 +6y

• (0,1)

= 0

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Graphical Check

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Graphical Check

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Graphical Check

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Algorithm.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable,

usually it’s x

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable,

usually it’s x, and the dependent variable

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable,

usually it’s x, and the dependent variable, that is, the function

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable,

usually it’s x, and the dependent variable, that is, the function, which is usually y.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable,

usually it’s x, and the dependent variable, that is, the function, which is usually y. To start with, it may be a good idea to replace “y” with “y(x)” or with “f(x)”.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable,

usually it’s x, and the dependent variable, that is, the function, which is usually y. To start with, it may be a good idea to replace “y” with “y(x)” or with “f(x)”.

• 2. Differentiate both sides of the equation.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable,

usually it’s x, and the dependent variable, that is, the function, which is usually y. To start with, it may be a good idea to replace “y” with “y(x)” or with “f(x)”.

• 2. Differentiate both sides of the equation.

Remember to use the chain rule for terms with y.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable,

usually it’s x, and the dependent variable, that is, the function, which is usually y. To start with, it may be a good idea to replace “y” with “y(x)” or with “f(x)”.

• 2. Differentiate both sides of the equation.

Remember to use the chain rule for terms with y.

• 3. Solve for y′.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable,

usually it’s x, and the dependent variable, that is, the function, which is usually y. To start with, it may be a good idea to replace “y” with “y(x)” or with “f(x)”.

• 2. Differentiate both sides of the equation.

Remember to use the chain rule for terms with y.

• 3. Solve for y′.

This is always possible, because, in any term that contains y′

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Algorithm. Computing a derivative by implicit

differentiation.

• 1. In the equation, determine the independent variable,

usually it’s x, and the dependent variable, that is, the function, which is usually y. To start with, it may be a good idea to replace “y” with “y(x)” or with “f(x)”.

• 2. Differentiate both sides of the equation.

Remember to use the chain rule for terms with y.

• 3. Solve for y′.

This is always possible, because, in any term that contains y′, the quantity y′ is just a factor.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1). Dependent and independent variables.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1). Dependent and independent variables. Independent: x

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1). Dependent and independent variables. Independent: x, dependent: y.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1). Dependent and independent variables. Independent: x, dependent: y. Differentiation with respect to x.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1). Dependent and independent variables. Independent: x, dependent: y. Differentiation with respect to x. xy3 +xy+3x = 2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1). Dependent and independent variables. Independent: x, dependent: y. Differentiation with respect to x. xy3 +xy+3x = 2 d dx

• xy3 +xy+3x
• =

d dx (2)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1). Dependent and independent variables. Independent: x, dependent: y. Differentiation with respect to x. xy3 +xy+3x = 2 d dx

• xy3 +xy+3x
• =

d dx (2) y3 +x3y2y′

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1). Dependent and independent variables. Independent: x, dependent: y. Differentiation with respect to x. xy3 +xy+3x = 2 d dx

• xy3 +xy+3x
• =

d dx (2) y3 +x3y2y′ +y+xy′

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1). Dependent and independent variables. Independent: x, dependent: y. Differentiation with respect to x. xy3 +xy+3x = 2 d dx

• xy3 +xy+3x
• =

d dx (2) y3 +x3y2y′ +y+xy′ +3

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Find the equation of the tangent line of the graph

defined by the equation xy3 +xy+3x = 2 at the point (2,−1). Dependent and independent variables. Independent: x, dependent: y. Differentiation with respect to x. xy3 +xy+3x = 2 d dx

• xy3 +xy+3x
• =

d dx (2) y3 +x3y2y′ +y+xy′ +3 =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

• 3xy2 +x
• y′

= −

• y3 +y+3
• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

• 3xy2 +x
• y′

= −

• y3 +y+3
• y′

= −y3 +y+3 3xy2 +x

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

• 3xy2 +x
• y′

= −

• y3 +y+3
• y′

= −y3 +y+3 3xy2 +x y′

• (2,−1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

• 3xy2 +x
• y′

= −

• y3 +y+3
• y′

= −y3 +y+3 3xy2 +x y′

• (2,−1)

= −y3 +y+3 3xy2 +x

• (2,−1)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

• 3xy2 +x
• y′

= −

• y3 +y+3
• y′

= −y3 +y+3 3xy2 +x y′

• (2,−1)

= −y3 +y+3 3xy2 +x

• (2,−1)

= −(−1)3 +(−1)+3 3·2·(−1)2 +2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

• 3xy2 +x
• y′

= −

• y3 +y+3
• y′

= −y3 +y+3 3xy2 +x y′

• (2,−1)

= −y3 +y+3 3xy2 +x

• (2,−1)

= −(−1)3 +(−1)+3 3·2·(−1)2 +2 = −1 8

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

• 3xy2 +x
• y′

= −

• y3 +y+3
• y′

= −y3 +y+3 3xy2 +x y′

• (2,−1)

= −y3 +y+3 3xy2 +x

• (2,−1)

= −(−1)3 +(−1)+3 3·2·(−1)2 +2 = −1 8 Equation of the tangent line

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

• 3xy2 +x
• y′

= −

• y3 +y+3
• y′

= −y3 +y+3 3xy2 +x y′

• (2,−1)

= −y3 +y+3 3xy2 +x

• (2,−1)

= −(−1)3 +(−1)+3 3·2·(−1)2 +2 = −1 8 Equation of the tangent line: y = −1 8(x−2)−1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

• 3xy2 +x
• y′

= −

• y3 +y+3
• y′

= −y3 +y+3 3xy2 +x y′

• (2,−1)

= −y3 +y+3 3xy2 +x

• (2,−1)

= −(−1)3 +(−1)+3 3·2·(−1)2 +2 = −1 8 Equation of the tangent line: y = −1 8(x−2)−1 = −1 8x− 3 4.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Solving for y′. y3 +x3y2y′ +y+xy′ +3 =

• 3xy2 +x
• y′

= −

• y3 +y+3
• y′

= −y3 +y+3 3xy2 +x y′

• (2,−1)

= −y3 +y+3 3xy2 +x

• (2,−1)

= −(−1)3 +(−1)+3 3·2·(−1)2 +2 = −1 8 Equation of the tangent line: y = −1 8(x−2)−1 = −1 8x− 3 4.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Graphical Check

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Graphical Check

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Graphical Check

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall?

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall? Volume of a right circular cone

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall? Volume of a right circular cone: V =

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall? Volume of a right circular cone: V = 1 3

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall? Volume of a right circular cone: V = 1 3πr2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall? Volume of a right circular cone: V = 1 3πr2h.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall? Volume of a right circular cone: V = 1 3πr2h. Sides rise at a 45◦ angle

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall? Volume of a right circular cone: V = 1 3πr2h. Sides rise at a 45◦ angle ⇒ r = h.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall? Volume of a right circular cone: V = 1 3πr2h. Sides rise at a 45◦ angle ⇒ r = h. So the volume of this cone, in terms

• f its height, is

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Gravel that is dumped off a conveyor belt forms a

right circular cone whose sides rise at a 45◦ angle. If 2m3 of gravel are dumped onto the cone every second, how fast is the height of the cone rising when the cone is 10m tall? Volume of a right circular cone: V = 1 3πr2h. Sides rise at a 45◦ angle ⇒ r = h. So the volume of this cone, in terms

• f its height, is V(h) = 1

3πh3.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3 d dtV = d dt 1 3πh3

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3 d dtV = d dt 1 3πh3

• V′

= 1 3π3h2h′

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3 d dtV = d dt 1 3πh3

• V′

= 1 3π3h2h′ = πh2h′

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3 d dtV = d dt 1 3πh3

• V′

= 1 3π3h2h′ = πh2h′ h′ = V′ πh2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3 d dtV = d dt 1 3πh3

• V′

= 1 3π3h2h′ = πh2h′ h′ = V′ πh2 h′

• h=10m

=

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3 d dtV = d dt 1 3πh3

• V′

= 1 3π3h2h′ = πh2h′ h′ = V′ πh2 h′

• h=10m

= V′ πh2

• h=10m

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3 d dtV = d dt 1 3πh3

• V′

= 1 3π3h2h′ = πh2h′ h′ = V′ πh2 h′

• h=10m

= V′ πh2

• h=10m

= 2m3

s

π(10m)2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3 d dtV = d dt 1 3πh3

• V′

= 1 3π3h2h′ = πh2h′ h′ = V′ πh2 h′

• h=10m

= V′ πh2

• h=10m

= 2m3

s

π(10m)2 = 1 50π m s

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3 d dtV = d dt 1 3πh3

• V′

= 1 3π3h2h′ = πh2h′ h′ = V′ πh2 h′

• h=10m

= V′ πh2

• h=10m

= 2m3

s

π(10m)2 = 1 50π m s ≈ 0.0064m s

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

V = 1 3πh3 d dtV = d dt 1 3πh3

• V′

= 1 3π3h2h′ = πh2h′ h′ = V′ πh2 h′

• h=10m

= V′ πh2

• h=10m

= 2m3

s

π(10m)2 = 1 50π m s ≈ 0.0064m s

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

Example.

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x d dx

• tan(y)
• =

d dxx

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x d dx

• tan(y)
• =

d dxx 1 cos2(y)y′ = 1

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x d dx

• tan(y)
• =

d dxx 1 cos2(y)y′ = 1 y′ = cos2(y)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x d dx

• tan(y)
• =

d dxx 1 cos2(y)y′ = 1 y′ = cos2(y) = cos2 arctan(x)

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x d dx

• tan(y)
• =

d dxx 1 cos2(y)y′ = 1 y′ = cos2(y) = cos2 arctan(x)

• =

cos2(y) cos2(y)+sin2(y)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x d dx

• tan(y)
• =

d dxx 1 cos2(y)y′ = 1 y′ = cos2(y) = cos2 arctan(x)

• =

cos2(y) cos2(y)+sin2(y) = 1

cos2(y)+sin2(y) cos2(y)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x d dx

• tan(y)
• =

d dxx 1 cos2(y)y′ = 1 y′ = cos2(y) = cos2 arctan(x)

• =

cos2(y) cos2(y)+sin2(y) = 1

cos2(y)+sin2(y) cos2(y)

= 1 1+tan2(y)

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x d dx

• tan(y)
• =

d dxx 1 cos2(y)y′ = 1 y′ = cos2(y) = cos2 arctan(x)

• =

cos2(y) cos2(y)+sin2(y) = 1

cos2(y)+sin2(y) cos2(y)

= 1 1+tan2(y) = 1 1+tan2 arctan(x)

• Bernd Schr¨
• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x d dx

• tan(y)
• =

d dxx 1 cos2(y)y′ = 1 y′ = cos2(y) = cos2 arctan(x)

• =

cos2(y) cos2(y)+sin2(y) = 1

cos2(y)+sin2(y) cos2(y)

= 1 1+tan2(y) = 1 1+tan2 arctan(x) = 1 1+x2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation

logo1 Introduction Derivatives Related Rates Inverse Functions

• Example. Compute the derivative of f(x) = arctan(x).

y = arctan(x) tan(y) = x d dx

• tan(y)
• =

d dxx 1 cos2(y)y′ = 1 y′ = cos2(y) = cos2 arctan(x)

• =

cos2(y) cos2(y)+sin2(y) = 1

cos2(y)+sin2(y) cos2(y)

= 1 1+tan2(y) = 1 1+tan2 arctan(x) = 1 1+x2

Bernd Schr¨

• der

Louisiana Tech University, College of Engineering and Science Implicit Differentiation