Question

Find the value of $\dfrac{dy}{dx}$, where $\cot \left( xy \right)+xy=y$.

Answer 1

For this problem we will first write the differentiation of each term in the given equation, and then find the differentiation values of each term and substitute them in the given equation. We have function like $f\left( g\left( x \right) \right)$, so we will use the formula $\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)$ to differentiate the term $\cot \left( xy \right)$. To find the differentiation of the term $xy$ we will use $uv$ formula i.e. ${{\left( uv \right)}^{'}}={{u}^{'}}v+{{v}^{'}}u$. After that we will add the differentiation values of $\cot \left( xy \right)$ , $xy$ and equate it to the differentiation of $y$ and then by simplifying we will get the value of $\dfrac{dy}{dx}$.

Complete step-by-step solution
Given that, $\cot \left( xy \right)+xy=y$
Differentiating the above equation with respect to $x$, we will get
$\begin{aligned} & \dfrac{d}{dx}\left[ \cot \left( xy \right)+xy \right]=\dfrac{dy}{dx} \\\ & \dfrac{d}{dx}\left[ \cot \left( xy \right) \right]+\dfrac{d}{dx}\left( xy \right)=\dfrac{dy}{dx}...\left( \text{i} \right) \\\ \end{aligned}$
Here we have the terms $\dfrac{d}{dx}\left[ \cot \left( xy \right) \right]$, $\dfrac{d}{dx}\left( xy \right)$. Now we are going to find the values of each term individually.
The value of term $\dfrac{d}{dx}\left[ \cot \left( xy \right) \right]$ is given by using the formula $\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)$ by substituting $f\left( g\left( x \right) \right)=\cot \left( xy \right)$, $g\left( x \right)=xy$, then we will get
$\dfrac{d}{dx}\left[ \cot \left( xy \right) \right]=\dfrac{d}{dx}\left( \cot xy \right).\dfrac{d}{dx}\left( xy \right)$
We know that $\dfrac{d}{dx}\left( \cot x \right)=-{{\csc }^{2}}x$, then
$\dfrac{d}{dx}\left[ \cot \left( xy \right) \right]=-{{\csc }^{2}}xy.\dfrac{d}{dx}\left( xy \right)....\left( \text{ii} \right)$
Now the value of $\dfrac{d}{dx}\left( xy \right)$ can be obtained by using the formula ${{\left( uv \right)}^{'}}={{u}^{'}}v+{{v}^{'}}u$ where $u=x$, $v=y$.
$\begin{aligned} & \therefore \dfrac{d}{dx}\left( xy \right)=y.\dfrac{dx}{dx}+x.\dfrac{dy}{dx} \\\ & \Rightarrow \dfrac{d}{dx}\left( xy \right)=y+x\dfrac{dy}{dx} \\\ \end{aligned}$
From the value of $\dfrac{d}{dx}\left( xy \right)$, the value of $\dfrac{d}{dx}\left[ \cot \left( xy \right) \right]$ is obtained using equation $\left( \text{ii} \right)$.
$\therefore \dfrac{d}{dx}\left[ \cot \left( xy \right) \right]=-{{\csc }^{2}}xy\left( y+x\dfrac{dy}{dx} \right)$
Now we have the values of both $\dfrac{d}{dx}\left[ \cot \left( xy \right) \right]$, $\dfrac{d}{dx}\left( xy \right)$. Substituting these values in equation $\left( \text{i} \right)$.
$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \cot \left( xy \right) \right]+\dfrac{d}{dx}\left( xy \right) \\\ & \Rightarrow \dfrac{dy}{dx}=-{{\csc }^{2}}xy\left( y+x\dfrac{dy}{dx} \right)+y+x\dfrac{dy}{dx} \\\ \end{aligned}$
Expand the term $-{{\csc }^{2}}xy\left( y+x\dfrac{dy}{dx} \right)$ by using multiplication distributive law, then we will get
$\dfrac{dy}{dx}=-y{{\csc }^{2}}xy-x{{\csc }^{2}}xy\dfrac{dy}{dx}+y+x\dfrac{dy}{dx}$
Rearranging the terms in the above equation, so that all the terms including the function $\dfrac{dy}{dx}$ are at one side and remaining terms at one side, then we will have
$y{{\csc }^{2}}xy-y=-x{{\csc }^{2}}xy\dfrac{dy}{dx}+x\dfrac{dy}{dx}-\dfrac{dy}{dx}$
Taking common $y$ from $y{{\csc }^{2}}xy-y$ and $\dfrac{dy}{dx}$ from $-x{{\csc }^{2}}xy\dfrac{dy}{dx}+x\dfrac{dy}{dx}-\dfrac{dy}{dx}$, then we will have
$y\left( {{\csc }^{2}}xy-1 \right)=\left( -x{{\csc }^{2}}xy+x-1 \right)\dfrac{dy}{dx}$
Again taking $x$ common from the term $-x{{\csc }^{2}}xy+x$, then we will get
$y\left( {{\csc }^{2}}xy-1 \right)=\left[ -1-x\left( {{\csc }^{2}}xy-1 \right) \right]\dfrac{dy}{dx}$
From the trigonometric identity ${{\csc }^{2}}x-{{\cot }^{2}}x=1$, we have ${{\csc }^{2}}xy-1={{\cot }^{2}}xy$, then we will get
$\begin{aligned} & y{{\cot }^{2}}xy=\left( -1-x{{\cot }^{2}}xy \right)\dfrac{dy}{dx} \\\ & \Rightarrow y{{\cot }^{2}}xy=-\left( 1+x{{\cot }^{2}}xy \right)\dfrac{dy}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=-\dfrac{y{{\cot }^{2}}xy}{1+x{{\cot }^{2}}xy} \\\ \end{aligned}$

Note: We can also find the value of $\dfrac{d}{dx}\left[ \cot \left( xy \right) \right]$ by using a substitution method. In this method we will substituting $xy=t$
Differentiating above value with respect to we have
$\begin{aligned} & \dfrac{d}{dx}\left( xy \right)=\dfrac{dt}{dx} \\\ & \Rightarrow x+y\dfrac{dy}{dx}=\dfrac{dt}{dx} \\\ \end{aligned}$
Now the value $\dfrac{d}{dx}\left[ \cot \left( xy \right) \right]$ is
$\begin{aligned} & \dfrac{d}{dx}\left[ \cot \left( xy \right) \right]=\dfrac{d}{dx}\left( \cot t \right) \\\ & \Rightarrow \dfrac{d}{dx}\left[ \cot \left( xy \right) \right]=-{{\csc }^{2}}t.\dfrac{dt}{dx} \\\ & \Rightarrow \dfrac{d}{dx}\left[ \cot \left( xy \right) \right]=-{{\csc }^{2}}\left( xy \right).\left( x+y\dfrac{dy}{dx} \right) \\\ \end{aligned}$
From both the methods we got the same value.