Question

How do you use implicit differentiation to find $\dfrac{{dy}}{{dx}}$ given ${x^3} + 3{x^2}y + {y^3} = 8$?

Answer 1

An implicit function is a function, written in terms of both dependent and independent variables, whereas an explicit function is a function which is represented in terms of an independent variable. Here, we will apply derivatives on both sides of the given equation, and also we know that, a derivative of a constant is always zero. Next, we will use the sum rule and the product rule, on the LHS of the equation. After solving this, and keeping all the $\dfrac{{dy}}{{dx}}$ terms on one side and remaining terms on the other side and evaluating this, we will get the final output.

Complete step by step answer:
Given that, $\;{x^3} + 3{x^2}y + {y^3} = 8$
Differential both the sides of the equation, we will get,
$\Rightarrow \dfrac{{dy}}{{dx}}\left( {\;{x^3} + 3{x^2}y + {y^3}} \right) = \dfrac{{dy}}{{dx}}\left( 8 \right)$
We know that, the derivative of a constant is zero, and so we will have,
$\Rightarrow \dfrac{{dy}}{{dx}}\left( {\;{x^3} + 3{x^2}y + {y^3}} \right) = 0$

Removing the brackets, we will get,
$\Rightarrow \;\dfrac{{dy}}{{dx}}{x^3} + \dfrac{{dy}}{{dx}}3{x^2}y + \dfrac{{dy}}{{dx}}{y^3} = 0$
Use the sum rule on the terms ${x^3}$ and${y^3}$, we will get,
$\Rightarrow \;3{x^2} + \dfrac{{dy}}{{dx}}3{x^2}y + 3{y^2}\dfrac{{dy}}{{dx}} = 0$
Use the product rule on the term $3{x^2}y$, we will get,
$\Rightarrow \;3{x^2} + 3{x^2}\dfrac{{dy}}{{dx}} + 6xy + 3{y^2}\dfrac{{dy}}{{dx}} = 0$
Rearrange all the terms, we will get,
$\Rightarrow \;3{x^2} + 6xy + 3{x^2}\dfrac{{dy}}{{dx}} + 3{y^2}\dfrac{{dy}}{{dx}} = 0$

Use transposition method and move all the terms without $\dfrac{{dy}}{{dx}}$ to RHS, we will get,
$\Rightarrow \;\dfrac{{dy}}{{dx}}\left( {3{x^2} + 3{y^2}} \right) = - 3{x^2} - 6xy$
$\Rightarrow \;\dfrac{{dy}}{{dx}} = \dfrac{{ - 3{x^2} - 6xy}}{{3{x^2} + 3{y^2}}}$
On evaluation this, we will get,
$\Rightarrow \;\dfrac{{dy}}{{dx}} = \dfrac{{ - (3{x^2} + 6xy)}}{{3{x^2} + 3{y^2}}}$
Taking 3 common from both the numerator and denominator, we will get,
$\Rightarrow \;\dfrac{{dy}}{{dx}} = \dfrac{{ - 3({x^2} + 2xy)}}{{3({x^2} + {y^2})}}$
$\Rightarrow \;\dfrac{{dy}}{{dx}} = \dfrac{{ - ({x^2} + 2xy)}}{{({x^2} + {y^2})}}$
$\therefore \;\dfrac{{dy}}{{dx}} = \dfrac{{ - {x^2} - 2xy}}{{{x^2} + {y^2}}}$

Hence, using the implicit differential, for the given equation ${x^3} + 3{x^2}y + {y^3}$ we get the value of $\;\dfrac{{dy}}{{dx}} = \dfrac{{ - {x^2} - 2xy}}{{{x^2} + {y^2}}}$.

Note: In the case of differentiation, an implicit function can be easily differentiated without rearranging the function and differentiating each term instead. For example, $y = 3x + 1$ is explicit where y is a dependent variable and is dependent on the independent variable x. Implicit differentiation allows us to find slopes of tangents to curves that are clearly not functions (they fail the vertical line test).