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Mathematics Question on Continuity and differentiability

If y=tan−1x2−1y =tan ^{-1} \sqrt {x^2-1} then the ratio d2ydx2:dydx\frac {d^2y}{dx^2}: \frac {dy}{dx}=_________

A

1+2x2x(x2+1)\frac {1+2x^2}{x(x^2+1)}

B

x(xn+1)(1−2x2)\frac {x(x^n+1)}{(1-2x^2)}

C

x(x2−1)(1+2x2)\frac {x(x^2 -1)}{(1+2x^2)}

D

1−2x2x(x2−1)\frac {1- 2x^2 }{x(x^2-1)}

Answer

1−2x2x(x2−1)\frac {1- 2x^2 }{x(x^2-1)}

Explanation

Solution

y=tan⁡−1x2−1y=\tan ^{-1} \sqrt{x^{2}-1} Put {x=sec⁡θ dx=sec⁡θ⋅tan⁡θdθ\begin{cases} x=\sec \theta \\\ d x=\sec \theta \cdot \tan \theta d \theta \end{cases} y=tan⁡−1sec⁡2θ−1=tan⁡−1(tan⁡θ)=θy=\tan ^{-1} \sqrt{\sec ^{2} \theta-1}=\tan ^{-1}(\tan \theta)=\theta =sec⁡−1x=\sec ^{-1} x ⇒dydx=ddx(sec⁡−1x)=1xx2−1\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{x \sqrt{x^{2}-1}} d2ydx2=1x⋅−121(x2−1)3/2(2x)−1x2⋅1x2−1\frac{d^{2} y}{d x^{2}} =\frac{1}{x} \cdot \frac{-1}{2} \frac{1}{\left(x^{2}-1\right)^{3 / 2}}(2 x)-\frac{1}{x^{2}} \cdot \frac{1}{\sqrt{x^{2}-1}} =−1(x2−1)3/2−1x2(x2−1)1/2=-\frac{1}{\left(x^{2}-1\right)^{3 / 2}}-\frac{1}{x^{2}\left(x^{2}-1\right)^{1 / 2}} =−1x2(x2−1)3/2(x2+x2−1)=-\frac{1}{x^{2}\left(x^{2}-1\right)^{3 / 2}}\left(x^{2}+x^{2}-1\right) =−(2x2−1)x2(x2−1)3/2=-\frac{\left(2 x^{2}-1\right)}{x^{2}\left(x^{2}-1\right)^{3 / 2}} Now, d2ydx2:dydx\frac{d^{2} y}{d x^{2}}: \frac{d y}{d x} =−(2x2−1)x2(x2−1)3/2:1x(x2−1)1/2=-\frac{\left(2 x^{2}-1\right)}{x^{2}\left(x^{2}-1\right)^{3 / 2}}: \frac{1}{x\left(x^{2}-1\right)^{1 / 2}} d2ydx2:dydx=(1−2x2):x(x2−1)\frac{d^{2} y}{d x^{2}}: \frac{d y}{d x}=\left(1-2 x^{2}\right): x\left(x^{2}-1\right) or (d2ydx2dydx)\left(\frac{\frac{d^{2} y}{d x^{2}}}{\frac{d y}{d x}}\right) =(1−2x2)x(x2−1)=\frac{\left(1-2 x^{2}\right)}{x\left(x^{2}-1\right)}