# Examples of Differentiation of Implicit Functions

Example: Find $\frac{{dy}}{{dx}}$, if the given implicit function is ${x^3} + {y^3} = xy$

We have the given implicit function
${x^3} + {y^3} = xy$

Differentiating with respect to $x$, we have
$\frac{d}{{dx}}{x^3} + \frac{d}{{dx}}{y^3} = \frac{d}{{dx}}\left( {xy} \right)$

Where $xy$ is the product of two variables and using the product of the derivative, we have
$\begin{gathered} 3{x^2} + 3{y^2}\frac{{dy}}{{dx}} = x\frac{{dy}}{{dx}} + y\frac{d}{{dx}}x \\ \Rightarrow 3{x^2} + 3{y^2}\frac{{dy}}{{dx}} = x\frac{{dy}}{{dx}} + y \\ \Rightarrow 3{y^2}\frac{{dy}}{{dx}} – x\frac{{dy}}{{dx}} = y – 3{x^2} \\ \Rightarrow \left( {3{y^2} – x} \right)\frac{{dy}}{{dx}} = y – 3{x^2} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{y – 3{x^2}}}{{3{y^2} – x}} \\ \end{gathered}$

Example: Find the derivative of the implicit function $\sqrt x – \sqrt y = \sqrt {xy}$

We have the given implicit function
$\sqrt x – \sqrt y = \sqrt {xy}$

Differentiating with respect to $x$, we have
$\begin{gathered} \frac{d}{{dx}}\sqrt x – \frac{d}{{dx}}\sqrt y = \frac{d}{{dx}}\sqrt {xy} \\ \Rightarrow \frac{1}{{2\sqrt x }} – \frac{1}{{2\sqrt y }}\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {xy} }}\frac{d}{{dx}}\left( {xy} \right) \\ \end{gathered}$

Where $xy$ is the product of two variables and using the product of the derivative, we have
$\begin{gathered} \frac{1}{{2\sqrt x }} – \frac{1}{{2\sqrt y }}\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {xy} }}\left( {x\frac{{dy}}{{dx}} + y\frac{d}{{dx}}x} \right) \\ \Rightarrow \frac{1}{{2\sqrt x }} – \frac{1}{{2\sqrt y }}\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt {xy} }}\left( {x\frac{{dy}}{{dx}} + y} \right) \\ \Rightarrow \frac{1}{{2\sqrt x }} – \frac{1}{{2\sqrt y }}\frac{{dy}}{{dx}} = \frac{x}{{2\sqrt {xy} }}\frac{{dy}}{{dx}} + \frac{y}{{2\sqrt {xy} }} \\ \Rightarrow \frac{1}{{2\sqrt x }} – \frac{1}{{2\sqrt y }}\frac{{dy}}{{dx}} = \frac{{\sqrt x }}{{2\sqrt y }}\frac{{dy}}{{dx}} + \frac{{\sqrt y }}{{2\sqrt x }} \\ \Rightarrow \frac{{\sqrt x }}{{2\sqrt y }}\frac{{dy}}{{dx}} + \frac{1}{{2\sqrt y }}\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }} – \frac{{\sqrt y }}{{2\sqrt x }} \\ \Rightarrow \left( {\frac{{\sqrt x }}{{2\sqrt y }} + \frac{1}{{2\sqrt y }}} \right)\frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x }} – \frac{{\sqrt y }}{{2\sqrt x }} \\ \Rightarrow \left( {\frac{{\sqrt x + 1}}{{2\sqrt y }}} \right)\frac{{dy}}{{dx}} = \frac{{1 – \sqrt y }}{{2\sqrt x }} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{1 – \sqrt y }}{{2\sqrt x }} \times \frac{{2\sqrt y }}{{\sqrt x + 1}} \\ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\sqrt y – y}}{{x + \sqrt x }} \\ \end{gathered}$