Question

Find the derivative of $\left( {{x^2}y} \right)$ with respect to x .

Answer 1

Hint:In the given problem, we are required to differentiate $\left( {{x^2}y} \right)$ with respect to x. Since, $\left( {{x^2}y} \right)$ is a product function, so we will have to apply product rule of differentiation in the process of differentiating $\left( {{x^2}y} \right)$ . Also derivatives of basic algebraic and trigonometric functions must be remembered thoroughly.

Complete step by step solution:
To find derivative of $\left( {{x^2}y} \right)$ with respect to $x$ , we have to find differentiate $\left( {{x^2}y} \right)$with respect to $x$. The product rule of differentiation involves differentiating a product of two or more functions.

So, Derivative of $\left( {{x^2}y} \right)$ with respect to $x$can be calculated as $\dfrac{d}{{dx}}\left( {{x^2}y} \right)$ .

Now, using the product rule of differentiation, we know that,$\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right) + g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right)$

So, Applying product rule to $\dfrac{d}{{dx}}\left( {{x^2}y} \right)$, we get,
$ = \left( {\dfrac{d}{{dx}}\left( {{x^2}} \right)} \right)y + {x^2}\dfrac{d}{{dx}}\left( y \right)$

Substituting the derivative of $\left( {\dfrac{d}{{dx}}\left( {{x^2}} \right)} \right)$ as $2x$, we get,
$ = 2xy + {x^2}\dfrac{d}{{dx}}\left( y \right)$

On further simplifying, we get,
$ = 2xy + {x^2}\dfrac{{dy}}{{dx}}$

So, the derivative of the $\left( {{x^2}y} \right)$ is $\left( {2xy + {x^2}\dfrac{{dy}}{{dx}}} \right)$ .

Note: The given problem may also be solved using the first principle of differentiation. The derivatives of basic trigonometric and algebraic functions must be learned by heart in order to find derivatives of complex composite functions using product rule and chain rule of differentiation. The product rule of differentiation involves differentiating a product of two or more functions and the chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.