How do you differentiate f(x)=x^2 * sin4x using the product rule?

Jan 7, 2016

$f ' \left(x\right) = 2 x \sin \left(4 x\right) + 4 {x}^{2} \cos \left(4 x\right)$

Explanation:

By the product rule, the derivative of $u \left(x\right) v \left(x\right)$ is $u ' \left(x\right) v \left(x\right) + u \left(x\right) v ' \left(x\right)$. Here, $u \left(x\right) = {x}^{2}$ and $v \left(x\right) = \sin \left(4 x\right)$ so $u ' \left(x\right) = 2 x$ and $v ' \left(x\right) = 4 \cos \left(4 x\right)$ by the chain rule.

We apply it on $f$, so $f ' \left(x\right) = 2 x \sin \left(4 x\right) + 4 {x}^{2} \cos \left(4 x\right)$.

Jan 7, 2016

$f ' \left(x\right) = 2 x \cdot \left(\sin \left(4 x\right) + 2 x \cos \left(4 x\right)\right)$

Explanation:

Given a $f \left(x\right) = h \left(x\right) \cdot g \left(x\right)$ the rule is:

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

in this case:

$h \left(x\right) = {x}^{2}$
$g \left(x\right) = \sin \left(4 x\right)$

look at $g \left(x\right)$ it is a composite function where the argoument is $4 \cdot x$

$g \left(x\right) = s \left(p \left(x\right)\right)$

then

$g ' \left(x\right) = s ' \left(p \left(x\right)\right) \cdot p ' \left(x\right)$

$\frac{d}{\mathrm{dx}} f \left(x\right) = \frac{d}{\mathrm{dx}} {x}^{2} \cdot \sin \left(4 x\right) + {x}^{2} \cdot \frac{d}{\mathrm{dx}} \sin \left(4 x\right) \cdot \frac{d}{\mathrm{dx}} 4 x =$

$\frac{d}{\mathrm{dx}} {x}^{2} \cdot \sin \left(4 x\right) + {x}^{2} \cdot \frac{d}{\mathrm{dx}} \sin \left(4 x\right) \cdot 4 \frac{d}{\mathrm{dx}} x =$

$= 2 \cdot x \cdot \sin \left(4 x\right) + {x}^{2} \cdot \cos \left(4 x\right) \cdot 4 \cdot 1 =$

$2 x \cdot \sin \left(4 x\right) + 4 {x}^{2} \cos \left(4 x\right) = 2 x \cdot \left(\sin \left(4 x\right) + 2 x \cos \left(4 x\right)\right)$