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Differentiate the following function w.r.t. \( x.\) $ (sin\: x)^{\large cos^{-1}x} $

1 Answer

Toolbox:
  • $\log m^n=n\log m$
  • $\cos^{-1}x=-\large\frac{1}{\sqrt{1-x^2}}$
Step 1:
$y=(\sin x)^{\large \cos^{-1}x}$
Take $\log$ on both sides,
$\log y=\cos^{-1}x.\log(\sin x)$
Differentiating with respect to $x$ on both sides we get,
Apply product rule,
$\cos^{-1}x=-\large\frac{1}{\sqrt{1-x^2}}$
$\large\frac{1}{y}\frac{dy}{dx}=$$\cos^{-1}x.\large\frac{1}{\sin x}$$.\cos x+\log(\sin x).\large\frac{-1}{\sqrt{1-x^2}}$
Step 2:
$\therefore \large\frac{dy}{dx}$$=y\big[\cos^{-1}x.\cot x-\large\frac{\log(\sin x)}{\sqrt{1-x^2}}\big]$
$\Rightarrow (\sin x)^{\large \cos^{-1}x}\big[\cos^{-1}x.\cot x-\large\frac{\log(\sin x)}{\sqrt{1-x^2}}\big]$
answered Oct 3, 2013 by sreemathi.v
 

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