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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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If \( y=x^{sin\: x}+(log\:x)^x,\: find\: \large\frac{dy}{dx}.\)

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Toolbox:
  • $\log m^{\large n}=n\log m$
  • $\large\frac{d}{dx}$$(\log x)=\large\frac{1}{x}$
  • $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Step 1:
$y=x^{\sin x}+(\log x)^x$
Let $u=x^{\large \sin x}$
Take $\log$ on both sides
$\log u=\sin x\log x$
Differentiating w.r.t $x$ we get,
$\large\frac{1}{u}\frac{du}{dx}$$=\sin x\large\frac{1}{x}$$+\log x.\cos x$
$\qquad=\large\frac{\sin x}{x}$$+\cos x.\log x$
$\therefore \large\frac{du}{dx}$$=u[\large\frac{\sin x}{x}$$+\cos x.\log x]$
$ \large\frac{du}{dx}$$=x^{\large\sin x}[\large\frac{\sin x}{x}$$+\cos x.\log x]$
Step 2:
Let $v=(\log x)^x$
Take $\log$ on both sides
$\log v=x\log(\log x)$
Differentiate w.r.t $x$ we get,
$\large\frac{1}{v}\frac{dv}{dx}$$=x.\large\frac{1}{\log x}.\frac{1}{x}$$+\log(\log x).1$
$\large\frac{dv}{dx}$$=v[\large\frac{1}{\log x}+$$\log(\log x)]$
$\large\frac{dv}{dx}$$=(\log x)^x[\large\frac{1}{\log x}+$$\log(\log x)]$
Step 3:
$\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
$\quad\;\;\;=x^{\large\sin x}[\large\frac{\sin x}{x}$$+\cos x.\log x]+(\log x)^x[\large\frac{1}{\log x}+$$\log(\log x)]$
answered Sep 26, 2013 by sreemathi.v
 

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