If y-sinlogex prove that dydx-1-y2x-

y=sin (log x ) dy dx =cos (log x ) × d dx (log x )=[√( 1 sin ^ 2 (log x ) #160; ) ] 1 x dy dx =(√( 1 y ^ 2 ) )× 1 x =√( 1 y ...

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Chain Rule of Differentiation

If y=sinlog...

Question

If $y = sin (l o g_{e} x)$ prove that $\frac{d y}{d x} = \frac{\sqrt{1 - y^{2}}}{x}$ .

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y = s i n (l o g_{e} x),

\frac{d y}{d x} = \frac{\sqrt{1 - y^{2}}}{x}

y = sin (2 {sin}^{- 1} x)

\frac{d y}{d x} = 2 \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}

x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}} = 1

\frac{d y}{d x} = - \sqrt{\frac{{1 - y}^{2}}{{1 - x}^{2}} .}

{cos}^{- 1} (\frac{x^{2} - y^{2}}{x^{2} + y^{2}}) = {tan}^{- 1} a

\frac{d y}{d x} = \frac{y}{x}

\sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = a (x - y)

\frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}

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