Question

Find the derivative of f(e^tanx) w r t x at x=0. f'(1) = 5

Answer 1

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SOLUTION:--


f (e^tanx) 
taking derivative wrt x we get 
f' (e^tanx) × [e^tanx] × [sec^2 (x)] 

putting x = 0 we get 
f'(e^0) ×[e^0] × [sec^2 (0) ] 
= f'(1) ×(1)×(1) 

It is given that f'(1) = 5 

》 Answer is 5 ×1×1 = 5

Answer 2

Answer:

Hence, the derlvative of$f\left(e^{\tan x}\right)$ wIth respect to x at x=0 Is 5 .

Step-by-step explanation:

We have function $f\left(e^{\tan x}\right)$ and given that$f^{\prime}(1)=5.$

Differentiate function $f\left(e^{\tan x}\right)$ with respect to x, we get

$\frac{d}{d x} f\left(e^{\tan x}\right)=f^{\prime}\left(e^{\tan x}\right) \frac{d}{d x} e^{\tan x} \text { (By chain rule) }$

$\Rightarrow \frac{d}{d x} f\left(e^{\tan x}\right)=f^{\prime}\left(e^{\tan x}\right) e^{\tan x} \frac{d}{d x} \tan x \text { (By chain rule) }$

$\Rightarrow \frac{d}{d x} f\left(e^{\tan x}\right)=f^{\prime}\left(e^{\tan x}\right) e^{\tan x} \sec ^{2} x \cdot\left(\because \frac{d}{d x} \tan x=\sec ^{2} x\right)$

Now, $\frac{d}{d x} f\left(e^{\tan x}\right)$ at $x=0$ is $\left[\frac{d}{d x} f\left(e^{\tan x}\right)\right]_{x=0}$

$=\left[f^{\prime}\left(e^{\tan x}\right) e^{\tan x} \sec ^{2} x\right]_{x=0}$

$=f^{\prime}\left(e^{\tan 0}\right) e^{\tan 0} \sec ^{2} 0=f^{\prime}(1) \times 1 \times 1=f^{\prime}(1)=5$

$\left.\left(\because \tan 0=0, \sec 0=1, e^{\tan 0}=e^{0}=1 \text { and } f^{\prime}(1)=5 \text { (given }$

Hence, the derlvative of$f\left(e^{\tan x}\right)$ wIth respect to x at x=0 Is 5 .

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