If y-sin-1 x2- prove that 1-x2d2ydx2-xdydx-2-0-

Given, y=(sin^ 1 x)^2 Differentiating with respect to x, we get dydx=2 sin^ 1 x ×1√(1 x^2) ⇒√(1 x^2) dydx=2sin^ 1 x Again, differentiating with respect to x, we ...

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Standard XII

Mathematics

Derivatives

If y=sin-1 x2...

Question

$If y = ({sin}^{- 1} x)^{2},$ prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - 2 = 0.$

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Solution

$Given, y = ({sin}^{- 1} x)^{2}$
Differentiating with respect to $x$ , we get
$\frac{d y}{d x} = 2 {sin}^{- 1} x \times \frac{1}{\sqrt{1 - x^{2}}}$
$\Rightarrow \sqrt{1 - x^{2}} \frac{d y}{d x} = 2 {sin}^{- 1} x$
Again, differentiating with respect to $x,$ we get
$\sqrt{1 - x^{2}} \frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} \times \frac{2 x}{2 \sqrt{1 - x^{2}}} = 2 \times \frac{1}{\sqrt{1 - x^{2}}}$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 2$
$\Rightarrow (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - 2 = 0$

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If y=(sin−1x)2, prove that (1−x2)d2ydx2−xdydx−2=0.

Given, y=(sin−1x)2 Differentiating with respect to x, we get dydx=2sin−1x×1√1−x2 ⇒√1−x2 dydx=2sin−1x Again, differentiating with respect to x, we get √1−x2 d2ydx2−dydx×2x2√1−x2=2×1√1−x2 ⇒(1−x2)d2ydx2−x dydx=2 ⇒(1−x2)d2ydx2−x dydx−2=0

$If y = ({sin}^{- 1} x)^{2},$ prove that $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - 2 = 0.$