Derivative of $\tan^{-1}(f(x))$

Question

What is derivative of $$\tan^{-1}\left(\frac{{\sqrt{4+x}+\sqrt{4-x}}}{\sqrt{4+x}-\sqrt{4-x}}\right).$$ So I tried to write it as $\tan(\tan^{-1}(...))$ to get the $f(x)=\frac{\pi}{4}+\tan^{-1}\left(\sqrt{\frac{4+x}{4-x}}\right)$ but still it's not better. Thanks help appreciated

Answer 1

HINT, using the chain rule and the quotient rule:

• $$\frac{\text{d}\arctan\left(f(x)\right)}{\text{d}x}=\frac{\frac{\text{d}f(x)}{\text{d}x}}{1+f(x)^2}=\frac{f'(x)}{1+f(x)^2}$$
• $$f'(x)=\frac{\text{d}}{\text{d}x}\left(\frac{2y(x)}{y(x)-v(x)}\right)=2\cdot\frac{(y(x)-v(x))\cdot\frac{\text{d}}{\text{d}x}\left(y(x)\right)-y(x)\cdot\frac{\text{d}}{\text{d}x}\left(y(x)-v(x)\right)}{\left(y(x)-v(x)\right)^2}$$

Answer 2

$$ \arctan(f(x)) = y \implies f(x) = \tan (y) $$ we want to find $y'$ thus we can perform $$ \frac{d}{dx}f(x) = \frac{d}{dx}\tan (y)= \sec^2 y \frac{dy}{dx} $$ so re-arranging $$ \frac{dy}{dx} = \cos^2 (y) \frac{df}{dx} $$ which is easier to solve.

Answer 3

Assuming the right expression to be $$\dfrac{1+\sqrt{\dfrac{4-x}{4+x}}}{1-\sqrt{\dfrac{4-x}{4+x}}}$$

$$\tan^{-1}\dfrac{1+\sqrt{\dfrac{4-x}{4+x}}}{1-\sqrt{\dfrac{4-x}{4+x}}}=\dfrac\pi4+\tan^{-1}\sqrt{\dfrac{4-x}{4+x}}$$

Let $\tan^{-1}\sqrt{\dfrac{4-x}{4+x}}=y\ge0\implies0\le y<\dfrac\pi2$

$$\dfrac{4-x}{4+x}=\tan^2y,\dfrac x4=\cdots=\cos2y\implies2y=\arccos\dfrac x4$$

Answer 4

$$\tan ^{ -1 } \left( \frac { \sqrt { 4+x } +\sqrt { 4-x } }{ \sqrt { 4+x } -\sqrt { 4-x } } \right) =\tan ^{ -1 } \left( \frac { { \left( \sqrt { 4+x } +\sqrt { 4-x } \right) }^{ 2 } }{ \left( \sqrt { 4+x } -\sqrt { 4-x } \right) \left( \sqrt { 4+x } +\sqrt { 4-x } \right) } \right) =\\ =\tan ^{ -1 }{ \left( \frac { 4+x-2\sqrt { 16-{ x }^{ 2 } } +4-x }{ 2x } \right) = } \tan ^{ -1 }{ \left( \frac { 8-2\sqrt { 16-{ x }^{ 2 } } }{ 2x } \right) } \\ \frac { d }{ dx } \left( \tan ^{ -1 }{ \left( \frac { 8-2\sqrt { 16-{ x }^{ 2 } } }{ 2x } \right) } \right) =\frac { 1 }{ 1+{ \left( \frac { 8-2\sqrt { 16-{ x }^{ 2 } } }{ 2x } \right) }^{ 2 } } \frac { \left( \frac { 2x }{ \sqrt { 16-{ x }^{ 2 } } } \right) 2x-2\left( 8-2\sqrt { 16-{ x }^{ 2 } } \right) }{ 4{ x }^{ 2 } } =\\ =\frac { 4{ x }^{ 2 } }{ 4{ x }^{ 2 }+{ \left( 8-2\sqrt { 16-{ x }^{ 2 } } \right) }^{ 2 } } \frac { 4{ x }^{ 2 }-16\sqrt { 16-{ x }^{ 2 } } +4\left( 16-{ x }^{ 2 } \right) }{ 4{ x }^{ 2 }\left( \sqrt { 16-{ x }^{ 2 } } \right) } =\frac { 4{ x }^{ 2 }-16\sqrt { 16-{ x }^{ 2 } } +4\left( 16-{ x }^{ 2 } \right) }{ \left( \sqrt { 16-{ x }^{ 2 } } \right) \left( 4{ x }^{ 2 }+{ \left( 8-2\sqrt { 16-{ x }^{ 2 } } \right) }^{ 2 } \right) } $$

Answer 5

Easy way :

Put $$x=4\cos 2\theta$$

Then $$y=\tan ^{-1} \left(\frac{\sqrt{4+4 \cos 2\theta}+\sqrt{4-4 \cos 2\theta}}{\sqrt{4+4 \cos 2\theta}-\sqrt{4-4 \cos 2\theta}} \right)$$

$$y=\tan ^{-1} \left(\frac{2\sqrt{1+ \cos 2\theta}+2\sqrt{1- \cos 2\theta}}{2\sqrt{1+ \cos 2\theta}-2\sqrt{1- \cos 2\theta}} \right)$$

$$y=\tan ^{-1} \left(\frac{\sqrt{1+ \cos 2\theta}+\sqrt{1- \cos 2\theta}}{\sqrt{1+ \cos 2\theta}-\sqrt{1- \cos 2\theta}} \right)$$

Since $1+\cos 2\theta =2\cos^2 \theta $ and $1-\cos 2\theta =2\sin^2 \theta$

$$y=\tan ^{-1} \left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta} \right)$$

Now divide both denominator and numerator by $\cos \theta$ ,

$$y=\tan ^{-1} \left(\frac{1+\tan \theta}{1-\tan \theta} \right)$$

$$y=\tan ^{-1} \left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4}\cdot\tan \theta} \right)$$

$$y=\tan ^{-1} \left(\tan\left( \frac{\pi}{4}-\theta \right)\right)$$

$$y=\frac{\pi}{4}-\theta$$

So $$\frac{dy}{dx}=0+\frac{d\theta}{dx}$$

Since $x=4 \cos 2\theta$

$$\frac{dx}{d \theta} = -8 \sin 2 \theta $$

So $$\frac{dy}{dx}=\frac{d\theta}{dx}= -\frac{1}{8\sin 2\theta}=-\frac{1}{8 \sqrt{1-\frac{x^2}{16}}}$$