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+1 vote

Differentiate w.r.t. \(x\) the function in \(( \log x )^{\large\log x}, x > 1 \)

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1 Answer

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Toolbox:
  • $(uv)'=u'v+uv'$
  • $\large\frac{dv}{dx}=\frac{dv}{dt}$$\times\large\frac{dt}{dx}$
Step 1:
Let $y=(\log x)^{\large\log x} \quad x > 1$
Taking $\log$ on both sides
$\log y=(\log x)\log(\log x)$
Here $u=\log x$
$v=\log(\log x)$
$\large\frac{du}{dx}=\frac{1}{x}$
Step 2:
$v=\log t$
$t=\log x$
$\large\frac{dv}{dt}=\frac{1}{t}$
$\large\frac{dt}{dx}=\frac{1}{x}$
$\large\frac{dv}{dx}=\frac{dv}{dt}$$\times\large\frac{dt}{dx}$
$\quad\;=\large\frac{1}{t}$$\times \large\frac{1}{x}$
$\quad\;=\large\frac{1}{\log x}$$\times \large\frac{1}{x}$
Step 3:
$(uv)'=u'v+uv'$
$\large\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}$$\log(\log x)+\log x.\large\frac{1}{\log x}.\large\frac{1}{x}$
$\large\frac{dy}{dx}$$=y[\large\frac{1}{x}$$\log(\log x)+\large\frac{1}{x}]$
$\quad\;=(\log x)^{\large\log x}[\large\frac{1}{x}+\frac{1}{x}$$\log\log x]$
$\Rightarrow \large\frac{1}{x}$$(\log x)^{\large\log x}[1+\log(\log x)]$
answered May 14, 2013 by sreemathi.v
 

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