Question

How do you find $\dfrac{{dy}}{{dx}}$ by implicit differentiation of ${x^3} - 3{x^2}y + 2x{y^2} = 10$ ?

Answer 1

Hint : Derivatives are defined as the varying rate of a function with respect to an independent variable. We know that implicit means some function of y and x equals something else. Knowing ‘x’ does not lead directly to ‘y’. we can solve this using product rule of differentiation, that is by $\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$ .

Complete step-by-step answer :
Given,
${x^3} - 3{x^2}y + 2x{y^2} = 10$
Now differentiate the above equation with respect to ‘x’.
$\Rightarrow \dfrac{d}{{dx}}\left( {{x^3} - 3{x^2}y + 2x{y^2}} \right) = \dfrac{d}{{dx}}\left( {10} \right)$
By linear combination rule, we have:
$\dfrac{{d({x^3})}}{{dx}} - \dfrac{{d(3{x^2}y)}}{{dx}} + \dfrac{{d(2x{y^2})}}{{dx}} = \dfrac{d}{{dx}}\left( {10} \right)$
Differentiation of constant is zero,
$\dfrac{{d({x^3})}}{{dx}} - \dfrac{{d(3{x^2}y)}}{{dx}} + \dfrac{{d(2x{y^2})}}{{dx}} = 0$
Applying differentiation for first term we have,
$3{x^2} - \dfrac{{d(3{x^2}y)}}{{dx}} + \dfrac{{d(2x{y^2})}}{{dx}} = 0$
Taking constant outside we have,
$3{x^2} - 3\dfrac{{d({x^2}y)}}{{dx}} + 2\dfrac{{d(x{y^2})}}{{dx}} = 0$
Now applying the product rule for the remaining two terms we have,
$3{x^2} - 3\left( {{x^2}\dfrac{{dy}}{{dx}} + y.2x} \right) + 2\left( {x\dfrac{{d({y^2})}}{{dx}} + {y^2}.1} \right) = 0$
$3{x^2} - 3\left( {{x^2}\dfrac{{dy}}{{dx}} + 2xy} \right) + 2\left( {x.2y\dfrac{{dy}}{{dx}} + {y^2}.1} \right) = 0$
$3{x^2} - 3\left( {{x^2}\dfrac{{dy}}{{dx}} + 2xy} \right) + 2\left( {2xy\dfrac{{dy}}{{dx}} + {y^2}} \right) = 0$
Expanding the brackets we have,
$3{x^2} - 3{x^2}\dfrac{{dy}}{{dx}} - 6xy + 4xy\dfrac{{dy}}{{dx}} + 2{y^2} = 0$
$- 3{x^2}\dfrac{{dy}}{{dx}} + 4xy\dfrac{{dy}}{{dx}} = 6xy - 3{x^2} - 2{y^2}$
Now taking $\dfrac{{dy}}{{dx}}$ common we have,
$\dfrac{{dy}}{{dx}}\left( { - 3{x^2} + 4xy} \right) = 6xy - 3{x^2} - 2{y^2}$
Now divide the whole differential equation by $\left( { - 3{x^2} + 4xy} \right)$ ,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{6xy - 3{x^2} - 2{y^2}}}{{ - 3{x^2} + 4xy}}$
So, the correct answer is “ $\dfrac{{dy}}{{dx}} = \dfrac{{6xy - 3{x^2} - 2{y^2}}}{{ - 3{x^2} + 4xy}}$ ”.

Note : $\bullet$ Linear combination rule: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as $h'(x) = af'(x) + bg'(x)$
$\bullet$ Quotient rule: The derivative of one function divided by other is found by
We know the differentiation of ${x^n}$ with respect to ‘x’ is $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$ . The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying product rules. We also know that differentiation of constant terms is zero.