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If \( y=(\log\: x)^x+(x)^{\large \cos\:x},\: find\: \large\frac{dy}{dx} \).

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  • $(uv)'=u'v+uv'$
  • $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Step 1:
$y=(\log x)^x+x^{\large\cos x}$
Let $u=(\log x)^x$
Take $\log$ on both sides,
$\log u=x\log (\log x)$
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}$$=x.\large\frac{1}{\log x}.\frac{1}{x}$$+\log (\log x).1$
$\qquad\quad=\large\frac{1}{\log x}$$+\log(\log x)$
$\large\frac{du}{dx}=$$u.[\large\frac{1}{\log x}$$+\log(\log x)]$
$\quad\;\;=$$(\log x)^x\large\frac{1}{\log x}$$+\log(\log x)$
Step 2:
Let $v=x^{\large\cos x}$
Take $\log$ on both sides,
$\log v=\cos x.\log x$
Differentiating with respect to $x$ on both sides we get,
$\large\frac{1}{v}\frac{dv}{dx}$$=\cos x.\large\frac{1}{x}$$+\log x(-\sin x)$
$\qquad=\large\frac{\cos x}{ x}$$-\sin x\log x$
$\large\frac{dv}{dx}=$$v.[\large\frac{\cos x}{ x}$$-\sin x\log x]$
$\large\frac{dv}{dx}=$$x^{\large\cos x}.[\large\frac{\cos x}{ x}$$-\sin x\log x]$
Step 3:
$\quad\;\;=(\log x)^x\large\frac{1}{\log x}$$+\log(\log x)+x^{\large\cos x}.[\large\frac{\cos x}{ x}$$-\sin x\log x]$
answered Oct 1, 2013 by sreemathi.v

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