Question

Differentiate y=$\tan \left\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\}$ with respect to $x$.

Answer 1

Hint: For solving this question we will use some standard results like differentiation of $y={{x}^{n}}$, $y=\tan x$ and then apply the chain rule of differentiation to differentiate the given term with respect to $x$ correctly.

Complete step-by-step answer:
Given:
We have to differentiate $y=\tan \left\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\}$ with respect to $x$.
Now, before we proceed we should be familiar with the following formulas and concepts of differential calculus:
1. If $y={{x}^{n}}$, then $\dfrac{dy}{dx}=n{{x}^{n-1}}$.
2. If $y=\tan x$, then $\dfrac{dy}{dx}={{\sec }^{2}}x$.
3. If $y=f\left\{ g\left( x \right) \right\}$, then $\dfrac{dy}{dx}={f}'\left\{ g\left( x \right) \right\}{g}'\left( x \right)$. This is also known as the chain rule of differentiation.
Now, we will use the above-mentioned formulas and concepts to differentiate $y=\tan \left\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\}$ with respect to $x$.
Now, let $y=\tan \left\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\}=f\left\{ g\left( x \right) \right\}$. Then,
  $f\left( x \right)=\tan x\text{ and }g\left( x \right)=\dfrac{\tan x-1}{\sqrt{\tan x}}={{\left( \tan x \right)}^{{}^{1}/{}_{2}}}-{{\left( \tan x \right)}^{-{}^{1}/{}_{2}}}$
Now, as $f\left( x \right)=\tan x$ so, we can write its differentiation with respect to $x$ from the formula written in the second point. Then,
$\begin{align}
  & f\left( x \right)=\tan x \\
 & \Rightarrow {f}'\left( x \right)={{\sec }^{2}}x \\
 & \Rightarrow {f}'\left\{ g\left( x \right) \right\}={{\sec }^{2}}\left\{ g\left( x \right) \right\} \\
 & \Rightarrow {f}'\left\{ g\left( x \right) \right\}={f}'\left\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\}={{\sec }^{2}}\left\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\}.................\left( 1 \right) \\
\end{align}$
Now, as $g\left( x \right)={{\left( \tan x \right)}^{{}^{1}/{}_{2}}}-{{\left( \tan x \right)}^{-{}^{1}/{}_{2}}}$ so, we can write its differentiation with respect to $x$ from the formula written in the above points. Then,
\[\begin{align}
  & g\left( x \right)={{\left( \tan x \right)}^{{}^{1}/{}_{2}}}-{{\left( \tan x \right)}^{-{}^{1}/{}_{2}}} \\
 & \Rightarrow {g}'\left( x \right)=\dfrac{1}{2}{{\left( \tan x \right)}^{\left( \dfrac{1}{2}-1 \right)}}\times {{\sec }^{2}}x-1\times \dfrac{-1}{2}{{\left( \tan x \right)}^{\left( -\dfrac{1}{2}-1 \right)}}\times {{\sec }^{2}}x \\
 & \Rightarrow {g}'\left( x \right)=\left[ \dfrac{{{\left( \tan x \right)}^{-\dfrac{1}{2}}}}{2}+\dfrac{{{\left( \tan x \right)}^{-\dfrac{3}{2}}}}{2} \right]{{\sec }^{2}}x \\
 & \Rightarrow {g}'\left( x \right)=\dfrac{{{\sec }^{2}}x{{\left( \tan x \right)}^{-\dfrac{3}{2}}}}{2}\left[ \tan x+1 \right]...................\left( 2 \right) \\
\end{align}\]
Now, as per our assumption $y=\tan \left\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\}=f\left\{ g\left( x \right) \right\}$ so, we can use the chain rule of differentiation to find $\dfrac{dy}{dx}$ . Then,
$\begin{align}
  & y=\tan \left\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\}=f\left\{ g\left( x \right) \right\} \\
 & \Rightarrow \dfrac{dy}{dx}={f}'\left\{ g\left( x \right) \right\}{g}'\left( x \right) \\
\end{align}$
Now, substituting expression of ${f}'\left\{ g\left( x \right) \right\}$ from equation (1) and \[{g}'\left( x \right)\] from equation (2) in the above equation. Then,
$\begin{align}
  & \dfrac{dy}{dx}={f}'\left\{ g\left( x \right) \right\}{g}'\left( x \right) \\
 & \Rightarrow \dfrac{dy}{dx}={{\sec }^{2}}\left\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\}\times \dfrac{{{\sec }^{2}}x{{\left( \tan x \right)}^{-\dfrac{3}{2}}}}{2}\left[ \tan x+1 \right] \\
\end{align}$
Thus, $\dfrac{dy}{dx}={{\sec }^{2}}\left\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\}\times \dfrac{{{\sec }^{2}}x{{\left( \tan x \right)}^{-\dfrac{3}{2}}}}{2}\left[ \tan x+1 \right]$ will be the differentiation of given function with respect to $x$.

Note: Here, the student should know how to apply the chain rule of differentiation to find the differentiation of functions of the form $y=f\left\{ g\left( x \right) \right\}$. Moreover, the student should proceed in a stepwise manner in the solution and avoid calculation mistakes while solving to get the correct answer.