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Differentiate the functions given in w.r.t. $x : $ $y=(\log x)^{\large x}+x^{\large\log x}$

This question has appeared in model paper 2012

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1 Answer

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Toolbox:
  • $(uv)'=u'v+uv'$
  • $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Step 1:
Let $y=(\log x)^{\large x}+x^{\large\log x}$
Now $u=(\log x)^{\large x}$
Taking $\log$ on both sides
$\log u=\log(\log x)^{\large x}$
$\qquad=x\log(\log x)$
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}=$$1.\log(\log x)+x.\large\frac{d}{dx}$$\log(\log x)$
$\qquad\;=\log(\log x)+x.\large\frac{1}{\log x}\large\frac{d}{dx}$$(\log x)$
$\qquad\;=\log(\log x)+x.\large\frac{1}{\log x}\large\frac{1}{ x}$
$\large\frac{du}{dx}$$=u[\log(\log x)+\large\frac{1}{\log x}]$
$\;\;\;\;=(\log x)^{\large x}$$[\log(\log x)+\large\frac{1}{\log x}]$
Step 2:
Now consider $v$
$v=x^{\large\log x}$
Taking $\log$ on both sides
$\log v=\log x^{\large\log x}$
$\qquad=\log x.\log x$
$\qquad=(\log x)^2$
Differentiating with respect to $x$
$\large\frac{1}{v}\frac{dv}{dx}=$$2\log x.\large\frac{d}{dx}$$(\log x)$
$\qquad\;=2\log x.\large\frac{1}{x}$
$\qquad\;=\large\frac{2}{x}$$\log x$
$\large\frac{dv}{dx}=$$v.[\large\frac{2}{x}$$\log x]$
Substitute the value of $v$
$\large\frac{dv}{dx}=$$x^{\large\log x}[\large\frac{2}{x}$$\log x]$
$\quad=\large\frac{2}{x}$$x^{\large\log x}$$\log x$
Step 3:
$\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
$\;\;\;\;=(\log x)^{\large x}$$[\log(\log x)+\large\frac{1}{\log x}]+\large\frac{2}{x}$$x^{\large\log x}$$\log x$
answered May 9, 2013 by sreemathi.v
 

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