# How do you differentiate (x^2) (sin x)?

Jul 4, 2016

By using the product rule.

#### Explanation:

Let $f \left(x\right) = \left({x}^{2}\right) \left(\sin x\right)$, then $f \left(x\right) = g \left(x\right) \times h \left(x\right)$.

The derivative of this function is given by $f ' \left(x\right) = \left(g ' \left(x\right) \times h \left(x\right)\right) + \left(h ' \left(x\right) \times g \left(x\right)\right)$

The derivative of $g \left(x\right)$ or ${x}^{2}$ is $g ' \left(x\right) = 2 \times {x}^{2 - 1} = 2 x$

The derivative of $h \left(x\right)$ or $\sin x$ is $h ' \left(x\right) = \cos x$.

Applying the product rule:

$f ' \left(x\right) = \left(g ' \left(x\right) \times h \left(x\right)\right) + \left(h ' \left(x\right) \times g \left(x\right)\right)$

$f ' \left(x\right) = \left(2 x \left(\sin x\right)\right) + \left({x}^{2} \left(\cos x\right)\right)$

$f ' \left(x\right) = 2 x \sin x + {x}^{2} \cos x$

Hence, the derivative of $y = \left({x}^{2}\right) \left(\sin x\right)$ is $y ' = 2 x \sin x + {x}^{2} \cos x$.

Hopefully this helps!