How do you differentiate f(x)=(x+3sinx)(x^2-3x) using the product rule?

Nov 2, 2016

$f ' \left(x\right) = \left(x + 3 \sin x\right) \left(2 x - 3\right) + \left({x}^{2} - 3 x\right) \left(1 + 3 \cos x\right)$

Explanation:

You need to use the product rule;
$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

So with $f \left(x\right) = \left(x + 3 \sin x\right) \left({x}^{2} - 3 x\right)$ we have;
$f ' \left(x\right) = \left(x + 3 \sin x\right) \left(\frac{d}{\mathrm{dx}} \left({x}^{2} - 3 x\right)\right) + \left({x}^{2} - 3 x\right) \left(\frac{d}{\mathrm{dx}} \left(x + 3 \sin x\right)\right)$

$\therefore f ' \left(x\right) = \left(x + 3 \sin x\right) \left(2 x - 3\right) + \left({x}^{2} - 3 x\right) \left(1 + 3 \cos x\right)$

We could multiply out, and it may possibly simplify but the above is probably sufficient as an answer.