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How do you do implicit differentiation for ${{x}^{2}}{{y}^{2}}+xy=2$?

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Last updated date: 26th Feb 2024
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IVSAT 2024
Answer
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Hint: For doing the implicit differentiation for the given equation ${{x}^{2}}{{y}^{2}}+xy=2$ we first have to choose our dependent and the independent variables. As per the usual convention, we choose the independent variable as $x$ and the dependent variable as $y$. Then we have to differentiate both sides of the given equation with respect to the independent variable $x$. From the differentiated expression, we have to separate $\dfrac{dy}{dx}$ in terms of the variables \[x\] and \[y\] to obtain the final derivative.

Complete step-by-step answer:
The equation given in the question is written as
${{x}^{2}}{{y}^{2}}+xy=2$
The implicit differentiation means finding out the derivative of the dependent variable with respect to the independent variable without expressing it explicitly in the form of the independent variable. So first we have to choose our dependent and the independent variables from the given equation. Let us choose $y$ as the dependent variable and $x$ as the independent variable. So we have to find out the derivative of $y$ with respect to $x$. For this, we differentiate both sides of the above equation with respect to $x$ to get
\[\begin{align}
  & \Rightarrow \dfrac{d\left( {{x}^{2}}{{y}^{2}}+xy \right)}{dx}=\dfrac{d\left( 2 \right)}{dx} \\
 & \Rightarrow \dfrac{d\left( {{x}^{2}}{{y}^{2}} \right)}{dx}+\dfrac{d\left( xy \right)}{dx}=\dfrac{d\left( 2 \right)}{dx} \\
\end{align}\]
We know that the differentiation of a constant is equal to zero. So the RHS of the above equation becomes zero and we get
$\Rightarrow \dfrac{d\left( {{x}^{2}}{{y}^{2}} \right)}{dx}+\dfrac{d\left( xy \right)}{dx}=0$
Now, from the product rule of differentiation we know that $\dfrac{d\left( fg \right)}{dx}=f\dfrac{d\left( g \right)}{dx}+g\dfrac{d\left( f \right)}{dx}$. So the above equation can be written as
\[\begin{align}
  & \Rightarrow {{x}^{2}}\dfrac{d\left( {{y}^{2}} \right)}{dx}+{{y}^{2}}\dfrac{d\left( {{x}^{2}} \right)}{dx}+x\dfrac{dy}{dx}+y\dfrac{dx}{dx}=0 \\
 & \Rightarrow {{x}^{2}}\dfrac{d\left( {{y}^{2}} \right)}{dx}+{{y}^{2}}\dfrac{d\left( {{x}^{2}} \right)}{dx}+x\dfrac{dy}{dx}+y=0 \\
\end{align}\]
From the chain rule of differentiation, we write the first term of the above equation as
\[\Rightarrow {{x}^{2}}\dfrac{d\left( {{y}^{2}} \right)}{dy}\left( \dfrac{dy}{dx} \right)+{{y}^{2}}\dfrac{d\left( {{x}^{2}} \right)}{dx}+x\dfrac{dy}{dx}+y=0\]
Now, we know that the differentiation of the function ${{x}^{n}}$ is equal to $n{{x}^{n-1}}$. So the above equation is written as
\[\begin{align}
  & \Rightarrow {{x}^{2}}\left( 2{{y}^{2-1}} \right)\dfrac{dy}{dx}+{{y}^{2}}\left( 2{{x}^{2-1}} \right)+x\dfrac{dy}{dx}+y=0 \\
 & \Rightarrow 2{{x}^{2}}y\dfrac{dy}{dx}+2x{{y}^{2}}+x\dfrac{dy}{dx}+y=0 \\
\end{align}\]
Writing the terms containing $\dfrac{dy}{dx}$ together, we have
\[\Rightarrow 2{{x}^{2}}y\dfrac{dy}{dx}+x\dfrac{dy}{dx}+2x{{y}^{2}}+y=0\]
Subtracting \[2x{{y}^{2}}+y\] from both the sides, we have
\[\begin{align}
  & \Rightarrow 2{{x}^{2}}y\dfrac{dy}{dx}+x\dfrac{dy}{dx}+2x{{y}^{2}}+y-\left( 2x{{y}^{2}}+y \right)=0-\left( 2x{{y}^{2}}+y \right) \\
 & \Rightarrow 2{{x}^{2}}y\dfrac{dy}{dx}+x\dfrac{dy}{dx}=-\left( 2x{{y}^{2}}+y \right) \\
\end{align}\]
Taking \[\dfrac{dy}{dx}\] common
\[\Rightarrow \left( 2{{x}^{2}}y+x \right)\dfrac{dy}{dx}=-\left( 2x{{y}^{2}}+y \right)\]
Finally, dividing both sides by \[\left( 2{{x}^{2}}y+x \right)\] we get
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{\left( 2x{{y}^{2}}+y \right)}{\left( 2{{x}^{2}}y+x \right)}\]
Now, taking out $y$ and $x$ common from the numerator and denominator respectively, we get
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{y\left( 2xy+1 \right)}{x\left( 2xy+1 \right)}\]
Finally cancelling \[\left( 2xy+1 \right)\], we get
$\Rightarrow \dfrac{dy}{dx}=-\dfrac{y}{x}$
Hence, this is the required implicit differentiation.

Note: It is not compulsory to choose $y$ as the dependent variable and $x$ as the independent variable. We can do the reverse too. Also, try to simplify the final expression for the derivative as much is possible. For example, in this case do not conclude \[\dfrac{dy}{dx}=-\dfrac{\left( 2x{{y}^{2}}+y \right)}{\left( 2{{x}^{2}}y+x \right)}\] as the final derivative, as the factor \[\left( 2xy+1 \right)\] is still common to the numerator and the denominator which can be cancelled.