# How do you use the product Rule to find the derivative of (2x)(-sinx) + (cosx)(2) - 2(cosx)?

Aug 11, 2015

First of all, notice how this cancels to give:

$\left(2 x\right) \left(- \sin x\right)$

Much easier now. Using the product rule, you have:

$f \left(x\right) = g \left(x\right) h \left(x\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = g \left(x\right) \frac{\mathrm{dh}}{\mathrm{dx}} + h \left(x\right) \frac{\mathrm{dg}}{\mathrm{dx}}$

So, you get:

$= \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left[- \sin x\right] + \left(- \sin x\right) \cdot \frac{d}{\mathrm{dx}} \left[2 x\right]$

$= \left(2 x\right) \cdot \left(- \cos x\right) + \left(- \sin x\right) \cdot 2$

$= \textcolor{b l u e}{- 2 x \cos x - 2 \sin x}$