Solveeit Logo

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 xx.

Explanation

Solution

Hint: For solving this question we will use some standard results like differentiation of y=xny={{x}^{n}}, y=tanxy=\tan x and then apply the chain rule of differentiation to differentiate the given term with respect to xx 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 xx.
Now, before we proceed we should be familiar with the following formulas and concepts of differential calculus:
1. If y=xny={{x}^{n}}, then dydx=nxn1\dfrac{dy}{dx}=n{{x}^{n-1}}.
2. If y=tanxy=\tan x, then dydx=sec2x\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 xx.
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)=tanx1tanx=(tanx)1/2(tanx)1/2f\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(x)=tanxf\left( x \right)=\tan x so, we can write its differentiation with respect to xx 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)1/2g\left( x \right)={{\left( \tan x \right)}^{{}^{1}/{}_{2}}}-{{\left( \tan x \right)}^{-{}^{1}/{}_{2}}} so, we can write its differentiation with respect to xx from the formula written in the above points. Then,

& 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{aligned}$$ Now, as per our assumption $y=\tan \left\\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\\}=f\left\\{ g\left( x \right) \right\\}$ so, we can use the chain rule of differentiation to find $\dfrac{dy}{dx}$ . Then, $\begin{aligned} & y=\tan \left\\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\\}=f\left\\{ g\left( x \right) \right\\} \\\ & \Rightarrow \dfrac{dy}{dx}={f}'\left\\{ g\left( x \right) \right\\}{g}'\left( x \right) \\\ \end{aligned}$ Now, substituting expression of ${f}'\left\\{ g\left( x \right) \right\\}$ from equation (1) and $${g}'\left( x \right)$$ from equation (2) in the above equation. Then, $\begin{aligned} & \dfrac{dy}{dx}={f}'\left\\{ g\left( x \right) \right\\}{g}'\left( x \right) \\\ & \Rightarrow \dfrac{dy}{dx}={{\sec }^{2}}\left\\{ \dfrac{\tan x-1}{\sqrt{\tan x}} \right\\}\times \dfrac{{{\sec }^{2}}x{{\left( \tan x \right)}^{-\dfrac{3}{2}}}}{2}\left[ \tan x+1 \right] \\\ \end{aligned}$ 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.