Question

Consider the following function, $y={{\left( \tan x \right)}^{\cot x}}+{{\left( \cot x \right)}^{\tan x}}$. Prove that \[\dfrac{dy}{dx}={{\left( \tan x \right)}^{\cot x}}.{{\csc }^{2}}x\left( 1-\log \tan x \right)+{{\left( \cot x \right)}^{\tan x}}.{{\sec }^{2}}x\left( \log \left( \cot x \right)-1 \right)\]

Answer 1

Hint: We solve this question by first dividing y into two parts and assuming them as $f\left( x \right)$ and $g\left( x \right)$ then differentiating $y$. Then we find the value of ${f}'\left( x \right)$ by first applying logarithm to $f\left( x \right)$ and then differentiating the function with respective to x. then we use the formulas for chain rule, $\dfrac{d}{dx}f\left( g\left( x \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$ and product rule, $\dfrac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right]={f}'\left( x \right)g\left( x \right)+f\left( x \right){g}'\left( x \right)$ and simplify it using the formulas, $\dfrac{d}{dx}\log x=\dfrac{1}{x}$, $\dfrac{d}{dx}\tan x={{\sec }^{2}}x$ and $\dfrac{d}{dx}\cot x=-{{\csc }^{2}}x$. Then by substituting the value of $f\left( x \right)$, we can find the value of ${f}'\left( x \right)$. Then we consider the function $g\left( x \right)$ and by following the same method we can find the value of ${g}'\left( x \right)$. Then we substitute the values of ${f}'\left( x \right)$ and ${g}'\left( x \right)$ to find the value of $\dfrac{dy}{dx}$.

Complete step by step answer:
Let us consider the given equation $y={{\left( \tan x \right)}^{\cot x}}+{{\left( \cot x \right)}^{\tan x}}$.
Let us assume that $f\left( x \right)={{\left( \tan x \right)}^{\cot x}}$ and $g\left( x \right)={{\left( \cot x \right)}^{\tan x}}$
So, now let us differentiate the given equation with respective to x. then we get,
$\Rightarrow \dfrac{dy}{dx}={f}'\left( x \right)+{g}'\left( x \right)...........\left( 1 \right)$

So, first let us consider $f\left( x \right)={{\left( \tan x \right)}^{\cot x}}$ and find the value of ${f}'\left( x \right)$.
$\Rightarrow f\left( x \right)={{\left( \tan x \right)}^{\cot x}}$
Now let us apply logarithm to the above function. Then we get,
$\Rightarrow \log f\left( x \right)=\log {{\left( \tan x \right)}^{\cot x}}$
Now let us consider the property $\log {{x}^{a}}=a\log x$.
Using this we can write the above equation as,
$\Rightarrow \log f\left( x \right)=\cot x.\log \left( \tan x \right)$
Now let us differentiate it with respective to x. Then we get,
$\Rightarrow \dfrac{d}{dx}\left( \log f\left( x \right) \right)=\dfrac{d}{dx}\left( \cot x.\log \left( \tan x \right) \right)..........\left( 2 \right)$
Let us consider the formula,
$\dfrac{d}{dx}\log x=\dfrac{1}{x}$
Now let us consider the formula for chain rule,
$\dfrac{d}{dx}f\left( g\left( x \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$
Using these formulas, we can write the left-hand side of the equation (2) as,
$\Rightarrow \dfrac{d}{dx}\left( \log f\left( x \right) \right)=\dfrac{1}{f\left( x \right)}\times {f}'\left( x \right).............\left( 3 \right)$

Let us consider the formula for the product rule,
$\dfrac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right]={f}'\left( x \right)g\left( x \right)+f\left( x \right){g}'\left( x \right)$
Now let us also consider the formulas,
$\begin{align}
  & \dfrac{d}{dx}\tan x={{\sec }^{2}}x \\
 & \dfrac{d}{dx}\cot x=-{{\csc }^{2}}x \\
\end{align}$
Let us consider the formula,
$\dfrac{d}{dx}\log x=\dfrac{1}{x}$
Now let us consider the formula for chain rule,
$\dfrac{d}{dx}f\left( g\left( x \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$

Using this we can write the right-hand side of the equation (2) as,
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( \cot x.\log \left( \tan x \right) \right)=\dfrac{d}{dx}\left( \cot x \right).\log \left( \tan x \right)+\cot x.\dfrac{d}{dx}\left( \log \left( \tan x \right) \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( \cot x.\log \left( \tan x \right) \right)=-{{\csc }^{2}}x.\log \left( \tan x \right)+\cot x.\left[ \dfrac{1}{\tan x}.{{\sec }^{2}}x \right] \\
 & \Rightarrow \dfrac{d}{dx}\left( \cot x.\log \left( \tan x \right) \right)=-{{\csc }^{2}}x.\log \left( \tan x \right)+{{\cot }^{2}}x.{{\sec }^{2}}x \\
\end{align}$
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( \cot x.\log \left( \tan x \right) \right)=-{{\csc }^{2}}x.\log \left( \tan x \right)+\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}.{{\sec }^{2}}x \\
 & \Rightarrow \dfrac{d}{dx}\left( \cot x.\log \left( \tan x \right) \right)=-{{\csc }^{2}}x.\log \left( \tan x \right)+{{\csc }^{2}}x \\
 & \Rightarrow \dfrac{d}{dx}\left( \cot x.\log \left( \tan x \right) \right)={{\csc }^{2}}x\left( 1-\log \left( \tan x \right) \right)...............\left( 4 \right) \\
\end{align}$
Substituting the equations (3) and (4) in equation (2), we get
$\begin{align}
  & \Rightarrow \dfrac{1}{f\left( x \right)}\times {f}'\left( x \right)={{\csc }^{2}}x\left( 1-\log \left( \tan x \right) \right) \\
 & \Rightarrow {f}'\left( x \right)=f\left( x \right)\times {{\csc }^{2}}x\left( 1-\log \left( \tan x \right) \right) \\
\end{align}$
Now let us substitute the value of $f\left( x \right)={{\left( \tan x \right)}^{\cot x}}$ in it. Then we get,
$\Rightarrow {f}'\left( x \right)={{\left( \tan x \right)}^{\cot x}}.{{\csc }^{2}}x\left( 1-\log \left( \tan x \right) \right).........\left( 5 \right)$

Now let us consider the second function, $g\left( x \right)={{\left( \cot x \right)}^{\tan x}}$.
$\Rightarrow g\left( x \right)={{\left( \cot x \right)}^{\tan x}}$
Now let us apply logarithm to the above function. Then we get,
$\Rightarrow \log g\left( x \right)=\log {{\left( \cot x \right)}^{\tan x}}$
Now let us consider the property $\log {{x}^{a}}=a\log x$.
Using this we can write the above equation as,
$\Rightarrow \log g\left( x \right)=\tan x.\log \left( \cot x \right)$
Now let us differentiate it with respective to x. Then we get,
$\Rightarrow \dfrac{d}{dx}\left( \log g\left( x \right) \right)=\dfrac{d}{dx}\left( \tan x.\log \left( \cot x \right) \right)..........\left( 6 \right)$
Let us consider the formula,
$\dfrac{d}{dx}\log x=\dfrac{1}{x}$
Now let us consider the formula for chain rule,
$\dfrac{d}{dx}f\left( g\left( x \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$
Using these formulas, we can write the left-hand side of the equation (6) as,
$\Rightarrow \dfrac{d}{dx}\left( \log g\left( x \right) \right)=\dfrac{1}{g\left( x \right)}\times {g}'\left( x \right).............\left( 7 \right)$

Let us consider the formula for the product rule,
$\dfrac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right]={f}'\left( x \right)g\left( x \right)+f\left( x \right){g}'\left( x \right)$
Now let us also consider the formulas,
$\begin{align}
  & \dfrac{d}{dx}\tan x={{\sec }^{2}}x \\
 & \dfrac{d}{dx}\cot x=-{{\csc }^{2}}x \\
\end{align}$
Let us consider the formula,
$\dfrac{d}{dx}\log x=\dfrac{1}{x}$
Now let us consider the formula for chain rule,
$\dfrac{d}{dx}f\left( g\left( x \right) \right)={f}'\left( g\left( x \right) \right)\times {g}'\left( x \right)$

Using this we can write the right-hand side of the equation (2) as,
$\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( \tan x.\log \left( \cot x \right) \right)=\dfrac{d}{dx}\left( \tan x \right).\log \left( \cot x \right)+\tan x.\dfrac{d}{dx}\left( \log \left( \cot x \right) \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( \tan x.\log \left( \cot x \right) \right)={{\sec }^{2}}x.\log \left( \cot x \right)+\tan x.\left[ \dfrac{1}{\cot x}.\left( -{{\csc }^{2}}x \right) \right] \\
 & \Rightarrow \dfrac{d}{dx}\left( \tan x.\log \left( \cot x \right) \right)={{\sec }^{2}}x.\log \left( \cot x \right)-{{\tan }^{2}}x.{{\csc }^{2}}x \\
\end{align}$
\[\begin{align}
  & \Rightarrow \dfrac{d}{dx}\left( \tan x.\log \left( \cot x \right) \right)={{\sec }^{2}}x.\log \left( \cot x \right)-\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}.{{\csc }^{2}}x \\
 & \Rightarrow \dfrac{d}{dx}\left( \tan x.\log \left( \cot x \right) \right)={{\sec }^{2}}x.\log \left( \cot x \right)-{{\sec }^{2}}x \\
 & \Rightarrow \dfrac{d}{dx}\left( \tan x.\log \left( \cot x \right) \right)={{\sec }^{2}}x\left( \log \left( \cot x \right)-1 \right)...............\left( 8 \right) \\
\end{align}\]
Substituting the equations (7) and (8) in equation (6), we get
$\begin{align}
  & \Rightarrow \dfrac{1}{g\left( x \right)}\times {g}'\left( x \right)={{\sec }^{2}}x\left( \log \left( \cot x \right)-1 \right) \\
 & \Rightarrow {g}'\left( x \right)=g\left( x \right)\times {{\sec }^{2}}x\left( \log \left( \cot x \right)-1 \right) \\
\end{align}$
Now let us substitute the value of $g\left( x \right)={{\left( \cot x \right)}^{\tan x}}$ in it. Then we get,
$\Rightarrow {g}'\left( x \right)={{\left( \cot x \right)}^{\tan x}}.{{\sec }^{2}}x\left( \log \left( \cot x \right)-1 \right).........\left( 9 \right)$

Now let us substitute the equations (5) and (9) in equation (1). Then we get,
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}={f}'\left( x \right)+{g}'\left( x \right) \\
 & \Rightarrow \dfrac{dy}{dx}={{\left( \tan x \right)}^{\cot x}}.{{\csc }^{2}}x\left( 1-\log \tan x \right)+{{\left( \cot x \right)}^{\tan x}}.{{\sec }^{2}}x\left( \log \left( \cot x \right)-1 \right) \\
\end{align}$
Hence Proved.

Note: The most commonly made mistake in this type of questions is one might take the formula for differentiation of the trigonometric identity $\cot x$ wrongly without taking the negative sign as, $\dfrac{d}{dx}\cot x={{\csc }^{2}}x$.