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Differentiate the functions given in w.r.t. $x : $ $ x^{\large x \cos x} + \large\frac{x^2 + 1}{x^2 - 1} $

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Toolbox:
  • $\big(\large\frac{u}{v}\big)'=\large\frac{u'v-uv'}{v^2}$
  • $\log m^{\large n}=n\log m$
Step 1:
Let $y=x^{\large\cos x}+\large\frac{x^2+1}{x^2-1}$
It is of the form $u+v$
Now $u=x^{\large x\cos x}$
Taking $\log $ on both sides
$\log u=\log x^{\large x\cos x}$
We know that $\log m^{\large n}=n\log m$
$\log u=x\cos x\log x$
Step 2:
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}=$$1.\cos x\log x+(-\sin x)(x\log x)+\large\frac{1}{x}$$(x\cos x)$
$[(uvw)'=u'vw+v'uw]$
$\Rightarrow \cos x\log x-x\sin x\log x+\cos x$
$\large\frac{du}{dx}$$=u[\cos x\log x-x\sin x\log x+\cos x]$
$\quad\;=x^{\large x\cos x}[\cos x\log x-x\sin x\log x+\cos x]$
Step 3:
Now consider $v$
$v=\large\frac{x^2+1}{x^2-1}$
$\large\frac{dv}{dx}=\frac{2x(x^2-1)-(x^2+1)2x}{(x^2-1)^2}$
$\quad=\large\frac{2x^3-2x-2x^3-2x}{(x^2-1)^2}$
$\quad=\large\frac{-4x}{(x^2-1)^2}$
Step 4:
$\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
$\quad\;=x^{\large x\cos x}[\cos x\log x-x\sin x\log x+\cos x]-\large\frac{4x}{(x^2-1)^2}$
answered May 9, 2013 by sreemathi.v
 

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