Question

How do you find the derivative of ${e^y} = x{y^2}$?

Answer 1

Derivatives are defined as the varying rate of a function with respect to an independent variable. To differentiate the right hand side of the equation we use the product rule. That is if we have $y = uv$ then its differentiation with respect to ‘x’ is $\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$. We solve this using implicit differentiation. We know that the differentiation of ${e^x}$ is ${e^x}$ with respect to ‘x’.

Complete step by step solution:
Given, ${e^y} = x{y^2}$
Now differentiate both sides with respect to ‘x’.
$\dfrac{d}{{dx}}({e^y}) = \dfrac{d}{{dx}}(x{y^2})$
Now applying the product rule$\dfrac{{d(uv)}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$
Where $u = x$ and $u = {y^2}$. Then we have
$\dfrac{d}{{dx}}({e^y}) = x \times \dfrac{{d({y^2})}}{{dx}} + {y^2} \times \dfrac{{d(x)}}{{dx}}$
Applying the differentiation
${e^y}\dfrac{{dy}}{{dx}} = x \times 2y\dfrac{{dy}}{{dx}} + {y^2} \times 1$
${e^y}\dfrac{{dy}}{{dx}} = 2xy\dfrac{{dy}}{{dx}} + {y^2}$
Grouping $\dfrac{{dy}}{{dx}}$ in on the left hand side of the equation,
${e^y}\dfrac{{dy}}{{dx}} - 2xy\dfrac{{dy}}{{dx}} = {y^2}$
Taking $\dfrac{{dy}}{{dx}}$ common, we have:
$\dfrac{{dy}}{{dx}}\left( {{e^y} - 2xy} \right) = {y^2}$
Dividing the whole equation by $\left( {{e^y} - 2xy} \right)$,
$\dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{\left( {{e^y} - 2xy} \right)}}$. This is the required answer.

Note: We know the differentiation of ${x^n}$ with respect to ‘x’ is $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$. We know the differentiation of ${y^n}$ with respect to ‘x’ is
$\dfrac{{d({y^n})}}{{dx}} = n.{y^{n - 1}}\dfrac{{dy}}{{dx}}$. We also have different rules in the differentiation. Those are
$\bullet$ Linear combination rules: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as $h'(x) = af'(x) + bg'(x)$

$\bullet$ Product rule: When a derivative of a product of two function is to be found, then we use product rule that is $\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$.
$\bullet$Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is $fog'({x_0}) = [(f'og)({x_0})]g'({x_0})$. We use these rules depending on the given problem.