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Find $\large \frac{dy}{dx},\normalsize if\;y=\large x^{tan x}+\sqrt{\large\frac{x^2+1}{2}}$

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  • If $y=[f(x)]^{g(x)}$,taking $\log$ on both sides we have $\log y=g(x).\log[f(x)]$.
Step 1:
$y=x^{\large\tan x}+\sqrt{\large\frac{x^2+1}{2}}$
Consider $y=x^{\large \tan x}$
Take $\log$ on both sides,
$\log y=\tan x.\log x$
Differentiating w.r.t $x$ on both sides we get,
$\large\frac{1}{y}\frac{dy}{dx}=$$\tan x.\large\frac{1}{x}$$+\log x.\sec^2 x$
Therefore $\large\frac{dy}{dx}=x^{\large \tan x}\bigg[\large\frac{\tan x}{x}$$+\log x.\sec^2 x\bigg]$-----(1)
Step 2:
Consider $y=\sqrt{\large\frac{x^2+1}{2}}$
Square on both sides
$y^2=\large\frac{x^2+1}{2}$
$2y^2=x^2+1$
Now differentiate w.r.t $x$ on both sides,
$2y\large\frac{dy}{dx}=$$2x$
$\Rightarrow \large\frac{dy}{dx}=\frac{x}{y}$
Substituting for $y$ we get,
Therefore $\large\frac{dy}{dx}=\large\frac{x\sqrt 2}{\sqrt{x^2+1}}$-----(2)
Step 3:
Combining equ(1) and equ(2) we get,
$\large\frac{dy}{dx}=x^{\large \tan x}\bigg[\large\frac{\tan x}{x}$$+\log x.\sec^2x\bigg]+\large\frac{x\sqrt 2}{\sqrt{x^2+1}}$
answered Jul 4, 2013 by sreemathi.v
 

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