Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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PLTL, MAT 131 * f(x)=keR⇒f'(x)=0 f(x)=x⇒f'(x)=1 f(x)=x*→ f'(x)= kxx-1 ƒ(x)=-=-=>ƒ'(x)=- — / 1 ƒ(x)=√x⇒ƒ'(x) = 2√x f(x)=lnx= f'(x)= Take a derivative. X f(x)=log x= f'(x)=- 1. f(x)=(x²+2x)√√x 2. f(x)=(x+1) 4x+7 3. 1 xlna √3x-1 -x²+3 f(x)=(x+1)(√x-1)(²+4) 4. f(x)= √x + 2x ²-1 5. f(x)=√√2x²+2x-1 6. f(x)= f(x)=e*f'(x)=e* f(x)=d⇒f'(x) = a* Ina f(x)=sinx⇒ f'(x) = cos x f(x)=cosx f'(x)=-sinx f(x)=tanx⇒ f'(x)=sec² x=1+tan² x 1 √1-x² 1 1+x² f(x)= arcsinx⇒ƒ'(x)=- f(x)= arctan x⇒f'(x)=
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Transcribed Image Text:PLTL, MAT 131 * f(x)=keR⇒f'(x)=0 f(x)=x⇒f'(x)=1 f(x)=x*→ f'(x)= kxx-1 ƒ(x)=-=-=>ƒ'(x)=- — / 1 ƒ(x)=√x⇒ƒ'(x) = 2√x f(x)=lnx= f'(x)= Take a derivative. X f(x)=log x= f'(x)=- 1. f(x)=(x²+2x)√√x 2. f(x)=(x+1) 4x+7 3. 1 xlna √3x-1 -x²+3 f(x)=(x+1)(√x-1)(²+4) 4. f(x)= √x + 2x ²-1 5. f(x)=√√2x²+2x-1 6. f(x)= f(x)=e*f'(x)=e* f(x)=d⇒f'(x) = a* Ina f(x)=sinx⇒ f'(x) = cos x f(x)=cosx f'(x)=-sinx f(x)=tanx⇒ f'(x)=sec² x=1+tan² x 1 √1-x² 1 1+x² f(x)= arcsinx⇒ƒ'(x)=- f(x)= arctan x⇒f'(x)=
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