Question

Here's Your "SwaggerGabru"

Harsh Pratap Singh :)

QUESTION :-
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Solve the problem in the attachment.

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Note :-
Spammed answer will be deleted ✔️
Giving copied answer is restricted ❎
Answer should be easy to understand
​

Answer 1

★FIND:

★FIND:

➡️log(f(e^x))

★GIVEN,

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

★SOLUTION:

Let,

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

Answer 2

$\tt \huge \red{answer : - } \\ \\ \tt \: given : \\ \star \sf \: f(1) = 4 \\ \star \sf \:f'(1) = 2 \\ \\ solution : - \\ \\ let \: \\ \sf \: y = log \: (f( {e}^{x} )) \\ \\ \sf \: \frac{dy}{dx} = \frac{dy}{dx} log \: (f( {e}^{x} )) \\ \\ \sf \: \frac{dy}{dx} = \frac{1}{f {(e}^{x} )} .f'( {e}^{x} ). {e}^{x} \\ \\ \sf \: \frac{dy}{dx} = \frac{f' {(e}^{x} ) }{f {(e}^{x} ). {e}^{x} } \\ \\ \sf \: \frac{dy}{dx} (x =0) = \frac{f'(e \degree)}{f(e \degree)} .e \degree = \frac{2}{4} = \frac{1}{2}$