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If y = cos inverse x, Find d ^2 y / dx ^2 in terms of y alone.

Q12  If  y = \cos ^{-1} x  Find   \frac{d ^2 y }{dx^2 }  in terms of y alone.

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Given function is
y = \cos ^{-1} x
Now, differentiation w.r.t. x
\frac{dy}{dx}=\frac{d( \cos ^{-1} x)}{dx}=\frac{-1}{\sqrt{1-x^2}}
Now, second order derivative is 
\frac{d^2y}{dx^2}=\frac{d^2(\frac{-1}{\sqrt{1-x^2}})}{dx^2}=\frac{-(-1)}{(\sqrt{1-x^2})^2}.(-2x) = \frac{-2x}{1-x^2}                                 -(i)
Now, we want \frac{d^2y}{dx^2} in terms of y
y = \cos ^{-1} x
x = \cos y
Now, put the value of x in (i)
\frac{d^2y}{dx^2} = \frac{-2\cos y }{1-\cos^2 y } = \frac{-2\cos y}{\sin ^2 y}= -2\cot y cosec y
                                                                                                    (\because 1-\cos^2x =\sin^2 x\ and \ \frac{\cos x}{\sin x} = \cot x \ and \ \frac{1}{\sin x}= cosec x)
Therefore, answer is  \frac{d^2y}{dx^2} = -2\cot y cosec y
 

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