Question

Find $\dfrac{dy}{dx}$ of $y=\tan {{x}^{\cot x}}+\cot {{x}^{\tan x}}$.

Answer 1

In this question, we will use sum rule $\dfrac{d}{dx}\left( a+b \right)=\dfrac{d}{dx}a+\dfrac{d}{dx}b$, product rule $\dfrac{d}{dx}\left( a.b \right)=b\dfrac{d}{dx}a+a\dfrac{d}{dx}b$. $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$,$\dfrac{d}{dx}\cot x=-{{\csc }^{2}}x$, $\dfrac{d}{dx}\log x=\dfrac{1}{x}dx$these are some basic trigonometric formulas used in this question.

Complete step by step answer:
From the question it is clear that we have to find $\dfrac{dy}{dx}$ of $y=\tan {{x}^{\cot x}}+\cot {{x}^{\tan x}}$.
$\Rightarrow y=\tan {{x}^{\cot x}}+\cot {{x}^{\tan x}}$…………(1)
Let us proceed to solve this question by taking $\tan {{x}^{\cot x}}=u$ and $\cot {{x}^{\tan x}}=v$ so that we can write $y=u+v$.
So, from sum rule, we can write
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( u+v \right)$
$\dfrac{dy}{dx}=\dfrac{d}{dx}u+\dfrac{d}{dx}v$…………(2)
Now consider $u=\tan {{x}^{\cot x}}$
$\Rightarrow u=\tan {{x}^{\cot x}}$
Apply log function on both sides. So, we get
$\Rightarrow \log u=\log \tan {{x}^{\cot x}}$
Now apply derivatives on both sides.
$\Rightarrow \dfrac{d}{dx}\log u=\dfrac{d}{dx}\left( log\tan {{x}^{\cot x}} \right)$
$\Rightarrow \dfrac{d}{dx}\log u=\dfrac{d}{dx}\left( \cot x\times log\tan x \right)$…………..(3)
We know that$\dfrac{d}{dx}\log x=\dfrac{1}{x}dx$, $\dfrac{d}{dx}\left( a.b \right)=b\dfrac{d}{dx}a+a\dfrac{d}{dx}b$. So, we can write equation (3) as
$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\cot x\times \dfrac{d}{dx}\left( logx\tan x \right)+\log x\tan x\times \dfrac{d}{dx}\cot x$
We know that $\dfrac{d}{dx}\cot x=-{{\csc }^{2}}x$, $\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$. So, we get equation as
$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\cot x\times \dfrac{1}{\tan x}\dfrac{d}{dx}\tan x+\log x\tan x\times \left( -{{\csc }^{2}}x \right)$
$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\cot x\times \dfrac{1}{\tan x}\left( {{\sec }^{2}}x \right)+\log x\tan x\times \left( -{{\csc }^{2}}x \right)$
From basic trigonometric reciprocal identities, we know $\dfrac{1}{\tan x}=\cot x$and $\cot x\times \sec x=\csc$ so
$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}=\cot x\times \cot x\left( {{\sec }^{2}}x \right)+\log x\tan x\times \left( -{{\csc }^{2}}x \right)$
$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}={{\cot }^{2}}x\left( {{\sec }^{2}}x \right)+\log x\tan x\times \left( -{{\csc }^{2}}x \right)$
$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}={{\left( \cot x\times \sec x \right)}^{2}}+\log x\tan x\times \left( -{{\csc }^{2}}x \right)$
$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}={{\csc }^{2}}x-\log x\tan x\times \left( {{\csc }^{2}}x \right)$
Take ${{\csc }^{2}}x$ common in RHS part
$\Rightarrow \dfrac{1}{u}\dfrac{du}{dx}={{\csc }^{2}}x\left( 1-\log x\tan x \right)$
Now multiply with $u$ on both sides.
$\Rightarrow u\times \dfrac{1}{u}\dfrac{du}{dx}=u\times {{\csc }^{2}}x\left( 1-\log x\tan x \right)$
After simplification we get,
$\Rightarrow \dfrac{du}{dx}=u\times {{\csc }^{2}}x\left( 1-\log x\tan x \right)$
Put $u=\tan {{x}^{\cot x}}$ in this equation
$\Rightarrow \dfrac{du}{dx}=\tan {{x}^{\cot x}}\times {{\csc }^{2}}x\left( 1-\log x\tan x \right)$……………….(4)
Now consider $v=\cot {{x}^{\tan x}}$
$\Rightarrow v=\cot {{x}^{\tan x}}$
Apply $\log$on both sides
$\Rightarrow \log v=\log \left( \cot {{x}^{\tan x}} \right)$
Differentiate on both sides, we get
$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\dfrac{d}{dx}\left( \tan x\log (\cot x) \right)$
$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\tan x\dfrac{1}{dx}\left( \log \cot x \right)+\log \left( \cot x \right)\dfrac{d}{dx}\tan x$
$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\tan x\times \dfrac{1}{\cot x}\dfrac{d}{dx}\cot x+\log \left( \cot x \right)\times {{\sec }^{2}}x$
$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=\tan x\times \dfrac{1}{\cot x}\left( -{{\csc }^{2}}x \right)+\log \left( \cot x \right)\times {{\sec }^{2}}x$
$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}={{\tan }^{2}}x\left( -{{\csc }^{2}}x \right)+\log \left( \cot x \right)\times {{\sec }^{2}}x$
$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}=-{{\sec }^{2}}x+\log \left( \cot x \right)\times {{\sec }^{2}}x$
In this equation take out ${{\sec }^{2}}x$common. So, we get
$\Rightarrow \dfrac{1}{v}\dfrac{dv}{dx}={{\sec }^{2}}x\left( -1+\log \cot x \right)$
Now multiply with $v$ on both sides. So, we get
$\Rightarrow v\times \dfrac{1}{v}\dfrac{dv}{dx}=v\times {{\sec }^{2}}x\left( -1+\log \cot x \right)$
After simplification we get,
$\Rightarrow \dfrac{dv}{dx}=v\times {{\sec }^{2}}x\left( -1+\log \cot x \right)$
Put $v=\cot {{x}^{\tan x}}$in this equation. So, we get
$\Rightarrow \dfrac{dv}{dx}=\cot {{x}^{\tan x}}\times {{\sec }^{2}}x\left( -1+\log \cot x \right)$…………….(5)
Put equation (4) and (5) in equation (2)
$\dfrac{dy}{dx}=\dfrac{d}{dx}u+\dfrac{d}{dx}v$
$\Rightarrow \dfrac{dy}{dx}=\tan {{x}^{\cot x}}\times {{\csc }^{2}}x\left( 1-\log x\tan x \right)+\cot {{x}^{\tan x}}\times {{\sec }^{2}}x\left( -1+\log \cot x \right)$
So finally, we got $\dfrac{dy}{dx}$
So, now we can conclude that $\tan {{x}^{\cot x}}\times {{\csc }^{2}}x\left( 1-\log x\tan x \right)+\cot {{x}^{\tan x}}\times {{\sec }^{2}}x\left( -1+\log \cot x \right)$ is the derivative of $\Rightarrow y=\tan {{x}^{\cot x}}+\cot {{x}^{\tan x}}$.

Note: students should be careful while doing calculations. Small error in the calculation may lead to this question being wrong. Students should know basic formulas and reciprocal identities in trigonometry. Using wrong formulas may lead to getting the wrong answer.