MATH 12002 - CALCULUS I 2.3, 2.4, and 2.5: Computing Derivatives - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 2.3, § 2.4, and § 2.5: Computing Derivatives (Part 2) Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 12

Derivatives 10 f ( x ) = sin( x 2 + 2) Use the Chain Rule: � d [cos( x 2 + 2)] · dx ( x 2 + 2) f ′ ( x ) � = [cos( x 2 + 2)] · 2 x = 2 x cos( x 2 + 2) . = D.L. White (Kent State University) 2 / 12

Derivatives 11 f ( x ) = tan(2 x − x 3 ) Use the Chain Rule: � d [sec 2 (2 x − x 3 )] · dx (2 x − x 3 ) f ′ ( x ) � = [sec 2 (2 x − x 3 )] · (2 − 3 x 2 ) = (2 − 3 x 2 ) sec 2 (2 x − x 3 ) . = D.L. White (Kent State University) 3 / 12

Derivatives 12 f ( x ) = x cos x Use the Product Rule: � d � d f ′ ( x ) � � = cos x + x · dx cos x dx x = 1 · cos x + x ( − sin x ) = cos x − x sin x . D.L. White (Kent State University) 4 / 12

Derivatives 13 f ( x ) = 4 sec 7 (2 − 4 x ) First rewrite the function as f ( x ) = 4[sec(2 − 4 x )] 7 , and then use the Chain Rule twice: � d 4 · 7[sec(2 − 4 x )] 6 · f ′ ( x ) � = dx sec(2 − 4 x ) � d 4 · 7[sec(2 − 4 x )] 6 · [sec(2 − 4 x ) tan(2 − 4 x )] · � = dx (2 − 4 x ) 4 · 7[sec(2 − 4 x )] 6 · [sec(2 − 4 x ) tan(2 − 4 x )] · ( − 4) = 28[sec 6 (2 − 4 x )] · [sec(2 − 4 x ) tan(2 − 4 x )] · ( − 4) . = D.L. White (Kent State University) 5 / 12

Derivatives 14 f ( x ) = (tan x )(sec x ) Use the Product Rule: � d � d f ′ ( x ) � � = dx tan x (sec x ) + (tan x ) dx sec x (sec 2 x )(sec x ) + (tan x )(sec x tan x ) . = D.L. White (Kent State University) 6 / 12

Derivatives 15 f ( x ) = cos 2 x − sin 2 x It may help to rewrite the function as f ( x ) = (cos x ) 2 − (sin x ) 2 . Use the Chain Rule on each term: 2(cos x ) 1 � d − 2(sin x ) 1 � d f ′ ( x ) � � = dx cos x dx sin x = 2(cos x )( − sin x ) − 2(sin x )(cos x ) = − 4(sin x )(cos x ) . Note: f ( x ) = cos 2 x − sin 2 x = 1 − 2 sin 2 x = 2 cos 2 x − 1. Try computing the derivative using these other two forms of f ( x ). D.L. White (Kent State University) 7 / 12

Derivatives 16 f ( x ) = 1 + tan x 1 − tan x Use the Quotient Rule: � d � d � � dx (1 + tan x ) (1 − tan x ) − (1 + tan x ) dx (1 − tan x ) f ′ ( x ) = (1 − tan x ) 2 (sec 2 x )(1 − tan x ) − (1 + tan x )( − sec 2 x ) = (1 − tan x ) 2 . D.L. White (Kent State University) 8 / 12

Derivatives 17 f ( x ) = sec x x 5 Use the Quotient Rule: � d � d · x 5 − (sec x ) · dx x 5 � � dx sec x f ′ ( x ) = ( x 5 ) 2 . (sec x tan x ) · x 5 − (sec x ) · 5 x 4 = ( x 5 ) 2 (sec x tan x ) · x 5 − (sec x ) · 5 x 4 = x 10 . D.L. White (Kent State University) 9 / 12

Derivatives 18 f ( x ) = sin( x 3 + 2) cos( x 3 + 2) Use the Quotient Rule, with the Chain Rule for the numerator and denominator: � d � d � � dx sin( x 3 +2) cos( x 3 +2) − [sin( x 3 +2)] dx cos( x 3 +2) f ′ ( x ) = [cos( x 3 +2)] 2 � d � d � � [cos( x 3 +2)] · dx ( x 3 +2) cos( x 3 +2) − [sin( x 3 +2)][ − sin( x 3 +2)] · dx ( x 3 +2) = [cos( x 3 +2)] 2 [cos( x 3 +2)] · 3 x 2 cos( x 3 +2) − [sin( x 3 +2)][ − sin( x 3 +2)] · 3 x 2 = [cos( x 3 +2)] 2 . Note: f ( x ) = sin( x 3 +2) cos( x 3 +2) = tan( x 3 + 2). Try computing the derivative using this other form of f ( x ). D.L. White (Kent State University) 10 / 12

Derivatives √ 19 f ( x ) = sin √ x + sin x √ Observe that sin √ x = sin( x 1 / 2 ) and sin x = (sin x ) 1 / 2 . Then use the Chain Rule on each term: (cos √ x ) · √ x � d � d 2 (sin x ) − 1 / 2 · + 1 � � f ′ ( x ) = dx sin x dx (cos √ x ) · 1 2 x − 1 / 2 + 1 2 (sin x ) − 1 / 2 cos x . = D.L. White (Kent State University) 11 / 12

Derivatives 20 f ( x ) = sin(tan x + sec x ) Use the Chain Rule: � d f ′ ( x ) � = [cos(tan x + sec x )] · dx (tan x + sec x ) [cos(tan x + sec x )](sec 2 x + sec x tan x ) . = D.L. White (Kent State University) 12 / 12