If y=\left ( \cot^{-1}x \right )^2 show that \left ( x^2+1 \right )^2\frac{\mathrm{d^2}y }{\mathrm{d} x^2}+2x\left (x^2+1 \right )\frac{\mathrm{d} y}{\mathrm{d} x}=2.

 

 

 

 
 
 
 
 

Answers (1)

\left ( x^2+1 \right )^2\frac{\mathrm{d^2}y }{\mathrm{d} x^2}+2x\left (x^2+1 \right )\frac{\mathrm{d} y}{\mathrm{d} x}=2

Differentiating wrt x

\frac{\mathrm{d} y}{\mathrm{d} x}=2\left ( \cot^{-1}x \right )\left ( \frac{-1}{1+x^2} \right )= \frac{-2\cot^{-1}x}{1+x^2}

Differentiating again

\frac{\mathrm{d^2} y}{\mathrm{d} x^2}=-2\left [ \frac{\left ( 1+x^2 \right )\left ( \frac{-1}{1+x^2} \right )-\cot^{-1}x(2x)}{\left (1+x^2 \right )^2} \right ]

\frac{\mathrm{d^2} y}{\mathrm{d} x^2}=2\left [ \frac{1+2x\cot^{-1}x}{(1+x^2)^2} \right ]

(1+x^2)^2\frac{\mathrm{d^2} y}{\mathrm{d} x^2}=2\left [ 1+2x\cot^{-1}x \right ]

\left ( x^2+1 \right )^2\frac{\mathrm{d^2}y }{\mathrm{d} x^2}+2x\left (x^2+1 \right )\frac{\mathrm{d} y}{\mathrm{d} x}=2

Hence Proved.

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