Question

How do you use implicit differentiation to find $\dfrac{{dy}}{{dx}}$ given$2{x^3} = {(3xy + 1)^2}$?

Answer 1

Implicit functions are differentiated with the help of chain rule to form an explicit function.
To use implicit differentiation we first differentiate both the sides with respect to $x$. Then we separate $\dfrac{{dy}}{{dx}}$to one side of the equation. Thus, we will solve the equation with respect to $\dfrac{{dy}}{{dx}}$.

Complete step by step answer:
The equation is as follows$2{x^3} = {(3xy + 1)^2}$
Differentiate both sides with respect to $x$
$\dfrac{{d}}{{dx}}\left( {2{x^3}} \right) = \dfrac{{d}}{{dx}}{\left( {3xy + 1} \right)^2}$
Now, we differentiate right hand side using chain rule.$\left( {\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}}.\dfrac{{du}}{{dx}}} \right)$
We get,
$3(2{x^2}) = 2(3xy + 1)\dfrac{d}{{dx}}(3xy + 1) \\\ 6{x^2} = 2(3xy + 1)\left( {\dfrac{d}{{dx}}(3xy) + \dfrac{d}{{dx}}(1)} \right) \\\ 6{x^2} = 2(3xy + 1)\left( {\dfrac{d}{{dx}}(3xy) + 0} \right) \\\ 6{x^2} = 2(3xy + 1)\dfrac{d}{{dx}}(3xy) \\\$
Now ,we use product rule to differentiate right hand side .$\left( {\dfrac{{d\left( {uv} \right)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}} \right)$
$6{x^2} = 2.3(3xy + 1)\dfrac{d}{{dx}}(xy) \\\ \not 6{x^2} = \not 6(3xy + 1)\left( {x\dfrac{{dy}}{{dx}} + y} \right) \\\ {x^2} = (3xy + 1)x\dfrac{{dy}}{{dx}} + y(3xy + 1) \\\ \dfrac{{{x^2} - y(3xy + 1)}}{{(3xy + 1)x}} = \dfrac{{dy}}{{dx}} \\\$
Hence, $\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} - y(3xy + 1)}}{{(3xy + 1)x}}$

Note: We use implicit differentiation as it’s not possible to reduce the implicit functions in a form where it is in terms of independent variables. If it’s an explicit function, then it can easily be presented in independent variables form and can also be differentiated easily using chain rule or other differentiation rule.