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Home  >>  CBSE XII  >>  Math  >>  Relations and Functions

Differentiate $x ^{\sin ^{-1} x}$ w.r.t $\sin^{-1} x$


$(A)\;x^{\sin^{-1}x}\begin{bmatrix} log\; x - \frac{\sqrt{1-x^2}}{x} \sin^{-1}x \end{bmatrix} $
$(B)\;x^{\sin^{-1}x}\begin{bmatrix} log\; x + \frac{\sqrt{1-x^2}}{x} \sin^{-1}x \end{bmatrix} $
$(C)\;x^{\sin^{-1}x}\begin{bmatrix}x\; log\; x + \frac{\sqrt{1-x^2}}{x} \sin^{-1}x \end{bmatrix} $
$(D)\;x^{\sin^{-1}x}\begin{bmatrix} x\;log\; x - \frac{\sqrt{1-x^2}}{x} \sin^{-1}x \end{bmatrix} $

1 Answer

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Let $u = \sin^{-1} x$ and $v=\sin^{-1}x$
$log\; u = \sin^{-1}x . log x$
$\implies \frac{1}{u} \frac{du}{dx} = \sin^{-1}x . \frac{1}{x} + log x . \frac{1}{\sqrt{1-x^2}}$
$\begin{align*} \implies \frac{du}{dx} &= u \begin{bmatrix} \frac{\sin^{-1} x}{x} + \frac{log x}{ \sqrt{1-x^2}} \end{bmatrix} \\ & = x^{sin^{-1} x} \begin{bmatrix} \frac{\sin^{-1}x}{x} + \frac{log \; x}{\sqrt{1-x^2}} \end{bmatrix} \end{align*}$
$v = \sin^{-1}x \implies \frac{dv}{dx} = \frac{1}{\sqrt{1-x^2}}$
$\begin{align*}\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{x^{\sin^{-1} x} \begin{bmatrix} \frac{\sin^{-1} x}{x} + \frac{log \; x }{\sqrt{1-x^2}}\end{bmatrix}}{\frac{1}{\sqrt{1-x^2}}} \end{align*}$
$ = x^{\sin^{-1}x}\begin{bmatrix} log\; x + \frac{\sqrt{1-x^2}}{x} \sin^{-1}x \end{bmatrix}$
answered Dec 21, 2016 by priyanka.c
 
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