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

1 Answer

  • $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
  • $\log m^{\large n}=n\log m$
  • According to product rule we have
  • $(uv)'=u'v+uv'$
Step 1:
Let $y=x^{\large\sin x}+(\sin x)^{\large\cos x}$
It is of the form
Let $u=x^{\large\sin x}$
Taking $\log$ on both sides
$\log u=\log x^{\large\sin x}$
$\qquad\;=\sin x\log x$
Step 2:
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}=$$\cos x\log x+\sin x$$.\large\frac{1}{x}$
We know that $(uv)'=u'v+uv'$
$\Rightarrow \cos x\log x+\large\frac{\sin x}{x}$
$\large\frac{du}{dx}=$$u.\cos x\log x$$.\large\frac{\sin x}{x}$
Substitute the value of $u$
$\Rightarrow x^{\large\sin x}(\cos x\log x+\large\frac{\sin x}{x})$
Step 3:
Consider $v=(\sin x)^{\large\cos x}$
Taking $\log$ on both sides
$\log v=\log(\sin x)^{\large\cos x}$
$\log m^{\large n}=n\log m$
$\Rightarrow \cos x\log(\sin x)$
Differentiating with respect to $x$
$\large\frac{1}{v}\frac{dv}{dx}$$=(-\sin x)\log(\sin x)+\cos x.\large\frac{d}{dx}$$\log(\sin x)$
We know that $(uv)'=u'v+uv'$
$u=\cos x$
$v=\log\sin x$
$\Rightarrow -\sin x\log(\sin x)+\cos x.\large\frac{1}{\sin x}.\frac{d}{dx}$$(\sin x)$
$\Rightarrow -\sin x\log(\sin x)+\large\frac{\cos x}{\sin x}.$$(\cos x)$
$\Rightarrow -\sin x\log(\sin x)+\large\frac{\cos^2 x}{\sin x}.$
$\large\frac{1}{v}\frac{dv}{dx}$$=(-\sin x)\log(\sin x)+\large\frac{\cos^2x}{\sin x}$
$\large\frac{dv}{dx}$$=v.[(-\sin x)\log(\sin x)+\large\frac{\cos^2x}{\sin x}]$
$\large\frac{dv}{dx}$$=(\sin x)^{\large\cos x}.[(-\sin x)\log(\sin x)+\large\frac{\cos^2x}{\sin x}]$
Step 4:
$\quad= x^{\large\sin x}(\cos x\log x+\large\frac{\sin x}{x})$+$(\sin x)^{\large\cos x}[\large\frac{\cos^2x-\sin^2x\log(\sin x)}{\sin x}]$
answered May 9, 2013 by sreemathi.v
edited May 9, 2013 by sreemathi.v