Question

How do you differentiate $xy=\cot \left( xy \right)$ ?

Answer 1

To differentiate $xy=\cot \left( xy \right)$, we are going to take the derivative with respect to x on both the sides. Now, the derivative is done using the product rule. In the product rule, two functions are written with a multiplication sign. Let us take two functions $f\left( x \right)\And g\left( x \right)$ and multiply them then we get $f\left( x \right)g\left( x \right)$. Now, applying the product rule on these two functions we get, $f\left( x \right)g'\left( x \right)+g\left( x \right)f'\left( x \right)$. The $'$ sign in the addition expression is the derivative sign.

Complete step by step answer:
The equation given in the above problem that we have to differentiate is as follows:
$xy=\cot \left( xy \right)$
Taking derivative with respect to x on both the sides using product rule which is equal to when two functions $f\left( x \right)\And g\left( x \right)$ are given in the following way:
$f\left( x \right)g\left( x \right)$
The derivative of the above two functions with respect to x is equal to:
$f\left( x \right)g'\left( x \right)+g\left( x \right)f'\left( x \right)$
Using the above product rule in $xy=\cot \left( xy \right)$ by taking derivative with respect to x on both the sides we get,
$xy=\cot \left( xy \right)$
We also know the derivative of $\cot x$ with respect to x is equal to $-\text{cose}{{\text{c}}^{2}}x$ so using this relation in finding the derivative of the R.H.S of the above equation we get,
$x\left( y' \right)+y\left( x' \right)=-\text{cose}{{\text{c}}^{2}}\left( xy \right)\left( x\left( y' \right)+y\left( x' \right) \right)$
You might have been thinking how we have taken the derivative of the R.H.S so we have first take the derivative of $\cot \left( xy \right)$ by assuming $xy$ as x then we have taken the derivative of $xy$ and multiplied with the result of the derivative of $\cot \left( xy \right)$ in which we have assumed $xy$ as x. This way of taking derivative is called chain rule.
Now, taking the derivative of the above equation we get,
$\begin{aligned} & x\left( \dfrac{dy}{dx} \right)+y\left( \dfrac{dx}{dx} \right)=-\text{cose}{{\text{c}}^{2}}\left( xy \right)\left( x\left( \dfrac{dy}{dx} \right)+y\left( \dfrac{dx}{dx} \right) \right) \\\ & \Rightarrow x\left( \dfrac{dy}{dx} \right)+y=-\text{cose}{{\text{c}}^{2}}\left( xy \right)\left( x\left( \dfrac{dy}{dx} \right)+y \right) \\\ & \Rightarrow x\left( \dfrac{dy}{dx} \right)+y=-x\left( \dfrac{dy}{dx} \right)\text{cose}{{\text{c}}^{2}}\left( xy \right)-y\text{cose}{{\text{c}}^{2}}\left( xy \right) \\\ \end{aligned}$
Now, writing $\dfrac{dy}{dx}$ terms on one side of the equation and the terms which do not contain $\dfrac{dy}{dx}$ on other side of the equation we get,
$x\dfrac{dy}{dx}+x\dfrac{dy}{dx}\text{cose}{{\text{c}}^{2}}\left( xy \right)=-y\text{cose}{{\text{c}}^{2}}\left( xy \right)-y$
In the above equation, taking $x\dfrac{dy}{dx}$ as common in the L.H.S and taking –y as common from R.H.S we get,
$x\dfrac{dy}{dx}\left( 1+\text{cose}{{\text{c}}^{2}}\left( xy \right) \right)=-y\left( 1+\text{cose}{{\text{c}}^{2}}\left( xy \right) \right)$
In the above, as you can see that $1+\text{cose}{{\text{c}}^{2}}\left( xy \right)$ is common in both the sides so we can cancel out this expression on both the sides and we get,
$x\dfrac{dy}{dx}=-y$
Dividing x on both the sides we get,
$\dfrac{dy}{dx}=-\dfrac{y}{x}$

Note: The possible mistake that could happen in this problem is missing the negative sign in the above solution. And also while taking the derivative of $\cot \left( xy \right)$ with respect to x you might forget to take the derivative of $xy$ with respect to x so make sure you have properly do the derivative of $\cot \left( xy \right)$ and be alert while doing the calculations.