Question

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

Answer 1

This problem deals with implicit differentiation of the given equation. Implicit differentiation is the procedure of differentiating an implicit equation with respect to the desired variable $x$ while treating the other variables as unspecified functions of $x$. To find $\dfrac{{dy}}{{dx}}$, we follow some procedures. Take $\dfrac{d}{{dx}}$ of both sides of the equation remembering to multiply by $\dfrac{{dy}}{{dx}}$ each time you see a $y$ term.

Complete step-by-step solution:
To implicitly derive a function. Differentiate with respect to $x$. Collect all the $\dfrac{{dy}}{{dx}}$ on one side. Solve for $\dfrac{{dy}}{{dx}}$.
Given the equation in terms of both $x$ and $y$, the equation is shown below:
$\Rightarrow {y^2} = 2 + xy$
Now differentiate the above equation on both sides with respect to $x$, on both sides of the equation, as given below:
$\Rightarrow \dfrac{d}{{dx}}\left( {{y^2}} \right) = \dfrac{d}{{dx}}\left( {2 + xy} \right)$
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( 2 \right) + x\dfrac{d}{{dx}}\left( y \right) + y\dfrac{d}{{dx}}\left( x \right)$
Here using the chain rule of differentiation, in the above expression, and as we know that the derivative of a constant is 0.
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = 0 + x\dfrac{{dy}}{{dx}} + y\left( 1 \right)$
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = x\dfrac{{dy}}{{dx}} + y$
Grouping the like terms and the unlike terms together as shown below;
$\Rightarrow 2y\dfrac{{dy}}{{dx}} - x\dfrac{{dy}}{{dx}} = y$
$\Rightarrow \left( {2y - x} \right)\dfrac{{dy}}{{dx}} = y$
Here taking the term $\dfrac{{dy}}{{dx}}$ is common in the left hand side of the equation.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{{\left( {2y - x} \right)}}$
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{y}{{2y - x}}$

The expression of $\dfrac{{dy}}{{dx}}$ is equal to $\dfrac{y}{{2y - x}}$ from implicit differentiation of given equation.

Note: Please note that the technique of implicit differentiation allows you to find the derivative of $y$ with respect to $x$ without having to solve the given equation for$y$. The chain rule must be used whenever the function $y$ is being differentiated because of our assumption that $y$ may be expressed as a function of $x$. The chain rule in differentiation is given by:
$\Rightarrow \dfrac{d}{{dx}}\left( {{f_1}(x).{f_2}(x)} \right) = {f_1}(x).\dfrac{d}{{dx}}\left( {{f_2}(x)} \right) + {f_2}(x).\dfrac{d}{{dx}}\left( {{f_1}(x)} \right)$
To derive an inverse function, restate it without the inverse then use implicit differentiation.