How do you differentiate f(x)= xsin3x using the product rule?

Feb 2, 2016

$\sin 3 x + 3 x \cos 3 x$

Explanation:

The product rule states that:
$\frac{d}{\mathrm{dx}} u \left(x\right) v \left(x\right) = u ' \left(x\right) v \left(x\right) + u \left(x\right) v ' \left(x\right)$

In our example, $u \left(x\right)$ (the first function) is simply $x$, while $v \left(x\right)$ (the other function) is $\sin 3 x$:
$x \cdot \sin 3 x$
$\textcolor{w h i t e}{\forall A}$
$u \left(x\right) \textcolor{w h i t e}{} v \left(x\right)$

Let's compute the derivatives of $x$ and $\sin 3 x$ here:
$x ' = 1$
$\left(\sin 3 x\right) ' = 3 \cos 3 x \to$(by the chain rule)

Now we plug everything into the chain rule equation:
$\frac{d}{\mathrm{dx}} x \sin 3 x = \left(x\right) ' \left(\sin 3 x\right) + \left(x\right) \left(\sin 3 x\right) '$
$\frac{d}{\mathrm{dx}} x \sin 3 x = \left(1\right) \left(\sin 3 x\right) + \left(x\right) \left(3 \cos 3 x\right)$
$\frac{d}{\mathrm{dx}} x \sin 3 x = \sin 3 x + 3 x \cos 3 x$