If y= ^ -1 x, Find d^2y dx^2 in terms of y alone.

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$\text{If y}=\cos^{-1}\text{x},\text{ Find }\frac{\text{d}^2\text{y}}{\text{dx}^2}\text{ in terms of y alone}.$

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Answer

$\text{y}=\cos^{-1}\text{x}\ \ \dots(1)$
$\therefore\ \frac{\text{dy}}{\text{dx}}=-\frac{1}{\sqrt{1-\text{x}^2}}=-(1-\text{x}^2)^{-\frac{1}{2}}$
$\therefore\ \frac{\text{dy}}{\text{dx}}=\frac{1}{2}(1-\text{x}^2)^{\frac{-3}{2}}(-2\text{x})=-\frac{\text{x}}{(1-\text{x}^2)^{\frac{3}{2}}}=-\frac{\cos\text{y}}{(1-\cos^2\text{y})^{\frac{3}{2}}}\ \ [\because\text{of }(1)]$
$=-\frac{\cos\text{y}}{(1-\cos^2\text{y})^{\frac{3}{2}}}=-\frac{\cos\text{y}}{\sin^3\text{y}}=-\frac{\cos\text{y}}{\sin\text{y}}.\frac{1}{\sin^2\text{y}}=-\cot\text{y cosec}^2\text{y}.$

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