Question

Differentiate ${{\tan }^{-1}}\dfrac{\sqrt{1+{{x}^{2}}}-1}{x}$ with respect to x.

Answer 1

Hint: We are going to use more than one substitution here. First we need to substitute $x=\tan \theta $ and simplify the differential as much as possible. We are going to use trigonometric ratios and formulas for sum of angles for sine and cosine as calculation requires.

Complete step by step answer:
Let us assume $x=\tan \theta $ where $\theta $ is any real number except $\dfrac{\pi }{2}+n\pi $ where $n$ is an integer.
\[\]Now we shall substitute $x=\tan \theta $ in the given differential. Let us assume
$y={{\tan }^{-1}}\dfrac{\sqrt{1+{{x}^{2}}}-1}{x}={{\tan }^{-1}}\dfrac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta }$ \[\]

Using the trigonometric formula ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$
$y={{\tan }^{-1}}\dfrac{\sqrt{1+{{x}^{2}}}-1}{x}={{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } \right)={{\tan }^{-1}}\dfrac{\sqrt{{{\sec }^{2}}\theta }-1}{\tan \theta }={{\tan }^{-1}}\left( \dfrac{\sec \theta -1}{\tan \theta } \right)$\[\]
Now we substitute the trigonometric ratio $\sec \theta =\dfrac{1}{\cos \theta }$ and $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ ,
$y={{\tan }^{-1}}\dfrac{\dfrac{1}{\cos \theta }-1}{\dfrac{\sin \theta }{\cos \theta }}={{\tan }^{-1}}\dfrac{1-\cos \theta }{\sin \theta }$ \[\]
We know from trigonometric formula that $\cos (A+B)=\cos A\cos B-\sin A\sin B$ and $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ where $A$ and $B$ are angles.
We substitute $A=B=\theta $ in the above cosine sum of angles formulas \[\]
$\begin{align}
  & \cos (\theta +\theta )=\cos \theta \cos \theta -\sin \theta \sin \theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\
 & \Rightarrow \cos (2\theta )={{\cos }^{2}}\theta -{{\sin }^{2}}\theta =1-{{\sin }^{2}}\theta -{{\sin }^{2}}\theta =1-2{{\sin }^{2}}\theta \\
 & \Rightarrow 2{{\sin }^{2}}\theta =1-\cos 2\theta \\
\end{align}$ \[\]
Replacing $\theta $ with $\dfrac{\theta }{2}$ we get $2{{\sin }^{2}}\theta =1-\cos \dfrac{\theta }{2}$\[\]
Similarly we substitute $A=B=\theta $ in the above sine sum of angles formulas,
$\begin{align}
  & \sin \left( \theta +\theta \right)=\sin \theta \cos \theta +\cos \theta \sin \theta =2\sin \theta \cos \theta \\
 & \Rightarrow \sin 2\theta =2\sin \theta \cos \theta \\
\end{align}$\[\]
Replacing $\theta $ with $\dfrac{\theta }{2}$ we get $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ \[\]
Using the trigonometric obtained above formula $ 1-\cos \theta =2{{\sin }^{2}}\left( \dfrac{\theta }{2} \right)$ and $\sin \theta =2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)$,\[y={{\tan }^{-1}}\left( \dfrac{1-\cos \theta }{\sin \theta } \right)={{\tan }^{-1}}\left( \dfrac{2{{\sin }^{2}}\left( \dfrac{\theta }{2} \right)}{\sin \theta =2\sin \left( \dfrac{\theta }{2} \right)\cos \left( \dfrac{\theta }{2} \right)} \right)={{\tan }^{-1}}\left( \tan \dfrac{\theta }{2} \right)=\dfrac{\theta }{2}\]
Replacing $\theta $ in the above result,
\[y=\dfrac{\theta }{2}=\dfrac{{{\tan }^{-1}}x}{2}\]
Differentiating above with respect to $x$ both side,
\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{{{\tan }^{-1}}x}{2} \right)=\dfrac{1}{2}\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{2\left( 1+{{x}^{2}} \right)}\]

The obtained value of the derivative is \[\dfrac{1}{2\left( 1+{{x}^{2}} \right)}\].

Note: Please note that the question is ambiguous about the value of $x$ in asked derivative because if $x=0$ we cannot differentiate. We need to take care of the domain and range of the function (in this case$\tan x$) during the calculation of inverse functions because sometimes the inverse for certain values may not even exist. In this case we have discarded the value $\dfrac{\pi }{2}+n\pi $ for any $\tan x$ from further calculation as at those values $\tan x$ is not defined.