# Differentiation of $y = \tan^{-1} \Bigl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\Bigr\}$

How do i differentiate the following: $$y = \tan^{-1} \biggl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\biggr\}$$

I know that $\text{derivative}$ of $\tan^{-1}{x}$ is $\frac{1}{1+x^{2}}$ but not sure as to how to do this.

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First the idea is to observe as to what would you substitute for $x$ in order to remove the square root. After some manipulations you find that the correct substitution is $x^{2} = \sin{2\theta}$. Once you have done this then you have $$y = \tan^{-1}\biggl\{ \frac{(\cos\theta + \sin\theta) - (\cos\theta - \sin\theta)}{(\cos\theta + \sin\theta)+(\cos\theta - \sin\theta)}\biggr\} = \theta = \frac{1}{2}\sin^{-1}{x^2}$$

So if $y = \frac{1}{2}\sin^{-1}{x^2}$ then $$\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1-x^{4}}} \cdot 2x = \frac{x}{\sqrt{1-x^{4}}}$$

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Your function simplifies to $\arctan \dfrac{1-\sqrt{1-x^4}}{x^2}$, which should be slightly easier to differentiate (remember the chain rule!)...

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Here are all the steps in excruciating detail.

Using the tangent of a sum formula gives $$\frac{1-\tan{\phi}}{1+\tan{\phi}}=\tan\left(\frac\pi4-\phi\right)\tag{1}$$ and letting $\tan(\phi)=\sqrt{\frac{1-x^2}{1+x^2}}$ yields \begin{align} \tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right) &=\tan^{-1}\left(\frac{1-\sqrt{\frac{1-x^2}{1+x^2}}}{1+\sqrt{\frac{1-x^2}{1+x^2}}}\right)\\ &=\frac\pi4-\tan^{-1}\left(\sqrt{\frac{1-x^2}{1+x^2}}\right)\tag{2} \end{align} Furthermore, we have \begin{align} \sqrt{\frac{1-\cos(\psi)}{1+\cos(\psi)}} &=\frac{\sin(\psi/2)}{\cos(\psi/2)}\\ &=\tan(\psi/2)\tag{3} \end{align} so letting $x^2=\cos(\psi)$ yields \begin{align} \tan^{-1}\left(\sqrt{\frac{1-x^2}{1-x^2}}\right) &=\psi/2\\ &=\frac12\cos^{-1}(x^2)\tag{4} \end{align} Combining $(2)$ and $(4)$ shows \begin{align} \tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right) &=\frac\pi4-\frac12\cos^{-1}(x^2)\\ &=\frac12\sin^{-1}(x^2)\tag{5} \end{align} Therefore, \begin{align} \frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}\left(\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}\right) &=\frac{\mathrm{d}}{\mathrm{d}x}\frac12\sin^{-1}(x^2)\\ &=\frac{x}{\sqrt{1-x^4}}\tag{6} \end{align}

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Putting $x^2=\cos2z,1+x^2=1+\cos2z=2\cos^2z,1-x^2=1-\cos2z=2\sin^2z$

$$\frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}=\frac{\cos z-\sin z}{\cos z+\sin z}=\frac{1-\tan z}{1+\tan z}=\tan\left(\frac\pi4-z\right)$$

$$y = \tan^{-1} \biggl\{ \frac{\sqrt{1+x^{2}} - \sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+ \sqrt{1-x^{2}}}\biggr\}=\tan^{-1}\tan\left(\frac\pi4-z\right)=\frac\pi4-z=\frac\pi4-\frac12\cos^{-1}x^2$$

$$\implies \frac{dy}{dx}=\frac{\frac\pi4-\frac12\cos^{-1}x^2}{dx}=-\frac12\left(-\frac{2x}{\sqrt{1-(x^2)^2}}\right)=\frac x{1-x^4}$$

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