How do you differentiate $({x^2})(\sin x)?$

Hint: In these types of questions, we need to use the formula of product rule which is given below as: $f'(x) = g'(x)h(x) + h'(x)g(x)$ . Find each and every term of the formula and put the values in it and get the solution as we require.

Formula used: 1) Product rule: $f'(x) = g'(x)h(x) + h'(x)g(x)$
2) $\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}$
3) $\dfrac{{d(\sin x)}}{{dx}} = \cos x$

Complete step-by-step solution:
We have given an expression as below:
We can consider the given function to be $f(x)$ .
Let $f(x) = ({x^2})(\sin x)$
Now, we need to define given function in the form of two new functions as $g\left( x \right)$ and $h\left( x \right)$
So, we get the following expression:
Then, $f(x) = g(x) \times h(x)$
The formula for the derivative of this function is product rule and it is mentioned below:
$f'(x) = (g'(x) \times h(x)) + (h'(x) \times g(x))......(A)$
The derivative of $g(x)$ or ${x^2}$ is equal to $g'(x) = 2 \times {x^{2 - 1}} = 2x.........(1)$ Using the formula $(2)$
The derivative of $h(x)$ or $\sin x$
$h'(x) = \cos x..........(2)$ …. Using the formula $(3)$
Applying the product rule:
$f'(x) = (g'(x) \times h(x)) + (h'(x) \times g(x))$
Putting the values we are getting in the equations $(1)$ and $(2)$ in the equation $(A)$ we can have,
$f'(x) = (2x(\sin x)) + ({x^2}(\cos x))$
Solve the brackets by multiplying the terms we get,
$f'(x) = 2x\sin x + {x^2}\cos x$
Hence, the derivative of $y = ({x^2})(\sin x)$
$\dfrac{{dy}}{{dx}} = $$y' = 2x\sin x + {x^2}\cos x$
Therefore, we differentiated the given function which we required.

Note: When finding the derivative of any function, in most cases, we have to use the product rule of differentiation. We have given the formula as below:
$f('x) = g'(x)h(x) + h'(x)g(x)$
This can be seen in the problem above where we had to use it.
-Differentiating is the method of finding the derivative of a function and finding the rate of change of function with respect to one.
-To differentiate something means to take the derivative of that value. Taking the derivative of a function is the same as finding the slope at any point, so differentiating is just similar to finding the slope.