Question

How do you differentiate $\left( {{x^2}} \right) - 2xy + {y^3}$?

Answer 1

For doing the implicit differentiation for the given function $\left( {{x^2}} \right) - 2xy + {y^3}$ we first have to choose our dependent and the independent variables. As per the usual convention, we choose the independent variable as $x$ and the dependent variable as $y$. Then we have to differentiate each term of the given function with respect to the independent variable $x$. We will use the power rule and product rule to solve the differentiation.
Formula used:
i). Chain Rule: Chain rule is applied when the given function is the function of function i.e.,
if y is a function of x, then $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$ or $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dv}} \times \dfrac{{dv}}{{dx}}$.
ii). The differentiation of the product of a constant and a function = the constant $\times$ differentiation of the function.
i.e., $\dfrac{d}{{dx}}\left( {kf\left( x \right)} \right) = k\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$, where $k$ is a constant.
iii). Power rule: $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}},n \ne - 1$
iv). Product rule: $\dfrac{d}{{dx}}\left( {fg} \right) = f\dfrac{d}{{dx

Complete step-by-step solution:
We have to differentiate $\left( {{x^2}} \right) - 2xy + {y^3}$.
The implicit differentiation means finding out the derivative of the dependent variable with respect to the independent variable without expressing it explicitly in the form of the independent variable. So first we have to choose our dependent and the independent variables from the given equation. Let us choose $y$ as the dependent variable and $x$ as the independent variable. So, we have to find out the derivative of $y$ with respect to $x$. For this, we differentiate both sides of the above equation with respect to $x$ , to get
$\dfrac{d}{{dx}}\left( {{x^2} - 2xy + {y^3}} \right)$
$\Rightarrow \dfrac{d}{{dx}}\left( {{x^2}} \right) - 2\dfrac{d}{{dx}}\left( {xy} \right) + \dfrac{d}{{dx}}\left( {{y^3}} \right)$
Now, from the product rule of differentiation, we know that $\dfrac{d}{{dx}}\left( {fg} \right) = f\dfrac{d}{{dx}}g + g\dfrac{d}{{dx}}f$. So, the above function can be written as
$\Rightarrow \dfrac{d}{{dx}}\left( {{x^2}} \right) - 2x\dfrac{d}{{dx}}\left( y \right) - 2y\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( {{y^3}} \right)$
From the chain rule of differentiation, we write the above function as
$\Rightarrow \dfrac{d}{{dx}}\left( {{x^2}} \right) - 2x\dfrac{d}{{dx}}\left( y \right) - 2y\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dy}}\left( {{y^3}} \right)\dfrac{{dy}}{{dx}}$
Now, we know that the differentiation of the function ${x^n}$ is equal to ${x^{n - 1}}$. So, the above function can be written as
$\Rightarrow 2x - 2x\dfrac{{dy}}{{dx}} - 2y + 3{y^2}\dfrac{{dy}}{{dx}}$
Take $2$ and $\dfrac{{dy}}{{dx}}$ common in above differentiation.
$\Rightarrow 2\left( {x - y} \right) + \left( {3{y^2} - 2x} \right)\dfrac{{dy}}{{dx}}$
Therefore, the differentiation of $\left( {{x^2}} \right) - 2xy + {y^3}$ is $2\left( {x - y} \right) + \left( {3{y^2} - 2x} \right)\dfrac{{dy}}{{dx}}$.

Note: The point to be noted here is that here in this question the given function is implicit function. So, we have to differentiate the equation using chain rule. The point to be remembered is that in implicit differentiation we have to differentiate each term. The differentiation of $y$ with respect to $x$ in implicit differentiation is $\dfrac{{dy}}{{dx}}$.