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

1 Answer

  • $\large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Step 1:
Let $y=(\sin x )^{\large x} + \sin^{-1} \sqrt x $
Its of the form
$\Rightarrow \large\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
Consider $u=(\sin x)^{\large x}$
Taking $\log$ on both sides
$\log u=\log(\sin x)^{\large x}$
We know that $\log m^{\large n}=n\log m$
Step 2:
Differentiating both sides with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}$$=1.\log \sin x+x.\large\frac{1}{\sin x}\large\frac{d}{dx}$$(\sin x)$
$\large\frac{1}{u}\frac{du}{dx}$$=\log \sin x+$$\large\frac{x\cos x}{\sin x}$
$\large\frac{du}{dx}$$=u\log \sin x+$$\large\frac{x\cos x}{\sin x}$
$\quad=(\sin x)^{\large x}$$[\log\sin x+x\cot x]$
Step 3:
Consider $v=\sin^{-1}\sqrt{x}$
Let $\sqrt x=t$
Put $v=\sin^{-1}t$
$\large\frac{dy}{dx}=\frac{dv}{dt}$$\times \large\frac{dt}{dx}$
$\quad\;=\large\frac{1}{\sqrt{1-t^2}}.\frac{1}{2\sqrt x}$
$\quad\;=\large\frac{1}{2\sqrt x\sqrt{1-x}}$
Step 4:
$\quad=(\sin x)^{\large x}$$[\log\sin x+x\cot x]$+$\large\frac{1}{2\sqrt{x-x^2}}$
answered May 9, 2013 by sreemathi.v

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