Question
Question: Differentiate y=\(\tan \left\\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\\}\) with respect to \(x\)....
Differentiate y=\tan \left\\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\\} with respect to x.
Solution
Hint: For solving this question we will use some standard results like differentiation of y=xn, y=tanx 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=xn, then dxdy=nxn−1.
2. If y=tanx, then dxdy=sec2x.
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(x)=tanx and g(x)=tanxtanx−1=(tanx)1/2−(tanx)−1/2
Now, as f(x)=tanx so, we can write its differentiation with respect to x from the formula written in the second point. Then,
\begin{aligned}
& 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{aligned}
Now, as g(x)=(tanx)1/2−(tanx)−1/2 so, we can write its differentiation with respect to x from the formula written in the above points. Then,