Question

find the derivative of (logx)^tanx​

Answer 1

Hii mate...

here is your answer...

step by step explainatio :-

y=(log x) ^ tan x

taking log both sides

we get,

log y = tan x log log x

differenciate both sides, we get

1/y (dy/dx)= (sec^2x log logx) +(tanx 1/logx 1/x)

dy/dx=y{(sec^2xloglogx)+(tanx 1/(xlogx))}

answer :-

dy /dx=[ (logs) ^ tanx ]{(sec^2xloglogx)+(tanx1/(xlogx)) }]

hope this will help you...

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Answer 2

Answer:

Correct answer is :

$\text{derivative\ =} [\sec^2x\cdot\log(\log x)+\frac{\tan x}{x\cdot\log x} ](\log x)^\tan x$

Step-by-step explanation:

Let assume that y = $(\log x)^\tan x$           ................( 1 ).

taking Log both sides in equation ( 1 )

                                       $\log y =\log (log x)^\tan x$

Differentiating above equation w.r.t to 'x' we get

$\frac{1}{y}\cdot\frac{dy}{dx}=\sec ^2 x\cdot \log (log x)+\frac{\tan x}{x\cdot\log x} \\ \frac{dy}{dx}=[\sec ^2 x\cdot \log (log x)+\frac{\tan x}{x\cdot\log x}]\cdot y \\\\\text{substituting the value of y we get}\\\\ \frac{dy}{dx}=[\sec ^2 x\cdot \log (log x)+\frac{\tan x}{x\cdot\log x}]\cdot (\log x)^\tan x$$\text{Hence derivative is =}[\sec ^2 x\cdot \log (log x)+\frac{\tan x}{x\cdot\log x}]\cdot (\log x)^\tan x$

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