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Differentiate the functions given in w.r.t. $x : $ $ (x + \large\frac{1}{x})^{\large x}$$ + x ^{\large (1 + \Large\frac{1}{x})} $

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Toolbox:
  • $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Step 1:
Let $y= (x + \large\frac{1}{x})^{\large x}$$ + x ^{\large (1 + \Large\frac{1}{x})} $
Let $u=(x+\large\frac{1}{x})^{\large x}$
$\;\;\;v=x^{\large(1+\Large\frac{1}{x})}$
$u=(x+\large\frac{1}{x})^{\large x}$
Taking $\log$ on both sides
$\log u=\log(x+\large\frac{1}{x})^{\large x}$
We know that $\log m^{\large n}=n\log m$
$\Rightarrow x\log(x+\large\frac{1}{x})$
$\Rightarrow x\log\large\frac{x^2+1}{x}$
$[\log\large\frac{m}{n}$$=\log m-\log n]$
$\Rightarrow x[\log (x^2+1)-\log x]$
Step 2:
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}=$$1.[\log(x^2+1)-\log x]+x[\large\frac{1}{x^2+1}\frac{d}{dx}$$(x^2+1)-\large\frac{1}{x}]$
$[(uv)'=u'v+uv']$
$\Rightarrow \log\large\frac{x^2+1}{x}$$+x\begin{bmatrix}\large\frac{1}{x^2+1}\normalsize \times 2x-\large\frac{1}{x}\end{bmatrix}$
$\Rightarrow \log\large\frac{x^2+1}{x}+\frac{2x^2}{x^2+1}$$-1$
$\Rightarrow \log(x+\large\frac{1}{x})+\large\frac{2x^2-x^2-1}{x^2+1}$
$\Rightarrow \log(x+\large\frac{1}{x})+\large\frac{x^2-1}{x^2+1}$
$\large\frac{du}{dx}$$=u\begin{bmatrix} \log(x+\large\frac{1}{x})+\large\frac{x^2-1}{x^2+1}\end{bmatrix}$
$\Rightarrow (x+\large\frac{1}{x})^x$$.\begin{bmatrix} \log(x+\large\frac{1}{x})+\large\frac{x^2-1}{x^2+1}\end{bmatrix}$
Step 3:
Consider $v=x^{\large(1+\Large\frac{1}{x})}$
We know that $\log m^{\large n}=n\log m$
$\Rightarrow v=(1+\large\frac{1}{x})$$\log x$
Differentiating with respect to $x$
$\large\frac{1}{v}\frac{dv}{dx}=\big(\frac{-1}{x^2}\big)$$\log x+\big(1+\large\frac{1}{x}\big)$$\times \large\frac{1}{x}$
$[(uv)'=u'v+uv']$
$\Rightarrow \large\frac{-1}{x^2}$$\log x+(1+\large\frac{1}{x})$$\times \large\frac{1}{x}$
$\Rightarrow \large\frac{-1}{x^2}$$\log x$$+\large\frac{x+1}{x^2}$
$\Rightarrow \large\frac{1}{x^2}$$(x+1-\log x)$
$\large\frac{dv}{dx}=\large\frac{v}{x^2}$$(x+1-\log x)$
Replacing the value of $v$ we get,
$\large\frac{dv}{dx}=\large\frac{x^{\Large(1+\Large\frac{1}{x})}}{x^2}$$(x+1-\log x)$
Step 4:
$\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
$\Rightarrow (x+\large\frac{1}{x})^x$$.\begin{bmatrix} \log(x+\large\frac{1}{x})+\large\frac{x^2-1}{x^2+1}\end{bmatrix}$+$\large\frac{x^{\Large(1+\Large\frac{1}{x})}}{x^2}$$(x+1-\log x)$
answered May 8, 2013 by sreemathi.v
 

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