MATH 12002 - CALCULUS I 5.4: Derivatives of Really General - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §5.4: Derivatives of Really General Exponential Functions

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 6

Even More General Exponential Functions

We have the differentiation formulas d dx xn = nxn−1 if n is constant and d dx ax = (ln a)ax if a is constant. So what is the derivative of f (x) = xx? Since the exponent is not constant,

d dx xx is NOT x · xx−1 (which is xx).

Since the base is not constant,

d dx xx is NOT (ln x)xx.

For a REALLY general exponential function of the form f (x)g(x), we use logarithmic differentiation.

D.L. White (Kent State University) 2 / 6

More Examples

Find the derivatives of the following functions:

1 f (x) = xx

First, we set y = xx, and then take natural logs of both sides: ln y = ln xx. Next, use the “Power” Law of Logarithms, ln(ar) = r ln a, to eliminate exponents: ln y = x ln x. Finally, take derivatives, with respect to x, of both sides: 1 y y′ = 1 · ln x + x · 1 x and then multiply both sides by y, recalling that y = xx: y′ = y(ln x + 1) = xx(ln x + 1). Hence f ′(x) = xx(ln x + 1).

D.L. White (Kent State University) 3 / 6

More Examples

2 f (x) = (x2 + 3)tan x

Again using logarithmic differentiation, we set y = (x2 + 3)tan x, and have ln y = ln

• (x2 + 3)tan x

= (tan x) ln(x2 + 3); 1 y y′ = (sec2 x) ln(x2 + 3) + (tan x) 1 x2 + 3 · 2x; y′ = y

• (sec2 x) ln(x2 + 3) + 2x tan x

x2 + 3

• .

Hence f ′(x) = (x2 + 3)tan x

• (sec2 x) ln(x2 + 3) + 2x tan x

x2 + 3

• .

D.L. White (Kent State University) 4 / 6

More Examples

Note: Logarithmic differentiation is necessary only if both the base and exponent are non-constant. It will work with normal exponential and power functions, using the fact that the derivative of a constant function is 0.

3 Use logarithmic differentiation to find the derivative of f (x) = 5x3+4.

Set y = 5x3+4, so that ln y = ln(5x3+4) = (x3 + 4) ln 5. Since ln 5 is just a constant, taking derivatives we get 1 y y′ = 3x2 ln 5, and so y′ = y · 3x2 ln 5. Hence f ′(x) = 3x2(ln 5)5x3+4. This is the same answer we would get using

d dx 5x = (ln 5)5x,

but requires much more work.

D.L. White (Kent State University) 5 / 6

More Examples

4 Use logarithmic differentiation to differentiate f (x) = (x3 + 4)5.

Set y = (x3 + 4)5, so that ln y = ln[(x3 + 4)5] = 5 ln(x3 + 4). Since 5 is just a constant, taking derivatives we get 1 y y′ = 5 · 1 x3 + 4 · 3x2, and so y′ = (x3 + 4)5 · 5 · 3x2 x3 + 4. Hence f ′(x) = 5(x3 + 4)4 · 3x2. This is the same answer we would get using

d dx x5 = 5x4,

but again requires much more work.

D.L. White (Kent State University) 6 / 6