Question

What is the derivative of \[{{\tan }^{-1}}\left( xy \right)=1+{{x}^{2}}y\]?

Answer 1

Hint: Here, the given question is to find the derivative of \[{{\tan }^{-1}}\left( xy \right)=1+{{x}^{2}}y\]. First, we differentiate both sides of the equation with respect to x. and by using chain rule, product rule and addition rule and also with some formulas, we are going to find out the derivative of the given question.

Complete step-by-step solution:
Let us solve the given question
Given that, \[{{\tan }^{-1}}\left( xy \right)=1+{{x}^{2}}y\]………………… (1)
On the left-hand side, we are going to use the chain rule in regards to the inverse tangent function
\[\dfrac{d}{dx}\left( \arctan \left( u \right) \right)=\dfrac{{{u}^{1}}}{1+{{u}^{2}}}\]…………………………… (2)
 Differentiating the both LHS and RHS
\[\Rightarrow \dfrac{d}{dx}\left( {{\tan }^{-1}}\left( xy \right) \right)=\dfrac{d}{dx}\left( 1+{{x}^{2}}y \right)\]
On LHS side, we used the formula which is equation (2) and on RHS side, we divided the terms on addition rule
\[\dfrac{\dfrac{d}{dx}\left( xy \right)}{1+{{\left( xy \right)}^{2}}}=\dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( {{x}^{2}}y \right)\]
We know that, derivative of any constant is zero,
\[\begin{align}
  & \dfrac{\dfrac{d}{dx}\left( xy \right)}{1+{{\left( xy \right)}^{2}}}=0+\dfrac{d}{dx}\left( {{x}^{2}}y \right) \\
 & \dfrac{\dfrac{d}{dx}\left( xy \right)}{1+{{\left( xy \right)}^{2}}}=\dfrac{d}{dx}\left( {{x}^{2}}y \right) \\
\end{align}\]
Here, we are using the product rule, to solve the above function
The formula for product rule is
\[\dfrac{d}{dx}\left( xy \right)=x\dfrac{dy}{dx}+y\]…………….. (3)
By using equation (3), on RHS, we get
\[\dfrac{\dfrac{d}{dx}\left( xy \right)}{1+{{\left( xy \right)}^{2}}}=y\dfrac{d}{dx}{{\left( x \right)}^{2}}+{{x}^{2}}\dfrac{d}{dx}\left( y \right)\]
We know that \[\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x\],
\[\Rightarrow \dfrac{y\dfrac{d}{dx}\left( x \right)+x\dfrac{d}{dx}\left( y \right)}{1+{{x}^{2}}{{y}^{2}}}=2xy+{{x}^{2}}\dfrac{dy}{dx}\]
\[\Rightarrow \dfrac{y+x\dfrac{dy}{dx}}{1+{{x}^{2}}{{y}^{2}}}=2xy+{{x}^{2}}\dfrac{dy}{dx}\]……………… (4)
Here, onwards we are trying to find out the \[\dfrac{dy}{dx}\]
 From equation (4), the LHS denominator term is sending to RHS, then
\[\Rightarrow y+x\dfrac{dy}{dx}=\left( 2xy+{{x}^{2}}\dfrac{dy}{dx} \right)\left( 1+{{x}^{2}}{{y}^{2}} \right)\]
On RHS, we are multiplying the terms, we get
 \[\Rightarrow y+x\dfrac{dy}{dx}=2xy\left( 1+{{x}^{2}}{{y}^{2}} \right)+\dfrac{dy}{dx}\left( {{x}^{2}}+{{x}^{4}}{{y}^{2}} \right)\]
Taking \[\dfrac{dy}{dx}\]terms on the left-hand side and other terms on the right hand side.
\[\Rightarrow x\dfrac{dy}{dx}-\dfrac{dy}{dx}\left( {{x}^{2}}+{{x}^{4}}{{y}^{2}} \right)=2xy+2{{x}^{3}}{{y}^{3}}-y\]
Taking \[\dfrac{dy}{dx}\] common on LHS, then
\[\Rightarrow \dfrac{dy}{dx}\left( x-{{x}^{2}}-{{x}^{4}}{{y}^{2}} \right)=2xy+2{{x}^{3}}{{y}^{3}}-y\]
\[\therefore\dfrac{dy}{dx}=\dfrac{2xy+2{{x}^{3}}{{y}^{3}}-y}{x-{{x}^{2}}-{{x}^{4}}{{y}^{2}}}\]

Note: A function in which the dependent and independent variables are not separated and the dependent variable is not on one side of the given equation is known as implicit function. The example for implicit function is our given question i.e., \[{{\tan }^{-1}}\left( xy \right)=1+{{x}^{2}}y\]. Here y is the independent variable and x is the independent variable, if we are differentiating the function with respect to x.