The derivative of ^-1(√1+x^2-1x) with respect to ^-1(2x√1-x^21-2x^2) at x=12 is:

Question

The derivative of $	an^{-1}\left(\dfrac{\sqrt{1+x^2}-1}{x}ight)$ with respect to $	an^{-1}\left(\dfrac{2x\sqrt{1-x^2}}{1-2x^2}ight)$ at $x=\dfrac12$ is:

Accepted Answer

Let $x=\tan\theta$. Then $u=\tan^{-1}\left(\dfrac{\sec\theta-1}{\tan\theta}\right)=\dfrac\theta2=\dfrac12\tan^{-1}x$, so $\dfrac{du}{dx}=\dfrac{1}{2(1+x^2)}$.

Let $x=\sin\phi$. Then $v=\tan^{-1}(\tan2\phi)=2\phi=2\sin^{-1}x$, so $\dfrac{dv}{dx}=\dfrac{2}{\sqrt{1-x^2}}$.

$$\frac{du}{dv}=\frac{1-x^2... }{}=\frac{\sqrt{1-x^2}}{4(1+x^2)}$$

At $x=\dfrac12$: $\dfrac{du}{dv}=\dfrac{\sqrt{3}/2}{4\cdot5/4}=\dfrac{\sqrt3}{10}$.

Solutions · 1