Take a derivative. 1. f(x)=(x²+2x)√x 2. f(x) = (x+1) 4x+7 3. f(x) = (x+1)(√x-D(²+4)

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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)=
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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