Math, asked by garvit1857, 1 year ago

if f(1)=4 and f'(1)=2 then find the value of the derivative log f (ex) wrt x at point x=0

Answers

Answered by MaheswariS

Answer:

The value of the derivative at x=0 is

$\bf\frac{1}{2}$

Step-by-step explanation:

Given:

f(1)=4 and f '(1)=2

$Let\;\;y=log(f(e^x))$

Differentiate with respect to x

By chain rule, we get

$\frac{dy}{dx}=\frac{1}{f(e^x)}\;f\,'(e^x)\;e^X$

At x=0,

$(\frac{dy}{dx})_{x=0}=\frac{1}{f(e^0)}\;f\,'(e^0)\;e^0$

$(\frac{dy}{dx})_{x=0}=\frac{1}{f(1)}\;f\,'(1)\1$

$(\frac{dy}{dx})_{x=0}=\frac{1}{4}\;(2)$

$(\frac{dy}{dx})_{x=0}=\frac{2}{4}$

$\implies\boxed{\bf(\frac{dy}{dx})_{x=0}=\frac{1}{2}}$

Answered by BRAINLYADDICTOR

$<marquee behaviour-move><h1>ANSWER:</ ht></marquee>$

NOw,

y=log(f(e^x))

dy/dx=dy/dx(log(f(e^x))

dy/dx=1/f(e^x).f'(e^x).e^x

dy/dx=f'(e^x)/f(e^x).e^x

dy/dx(x=0)=f'(e°)/f(e°).e°

=f'(1)/f(1).1

=2/4

=1/2

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