# How do you differentiate f(x) = (sinx)/(1-xe^x) using the quotient rule?

Jul 5, 2016

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

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(\frac{u \left(x\right)}{v \left(x\right)}\right) = \frac{u ' \left(x\right) \cdot v \left(x\right) - u \left(x\right) \cdot v ' \left(x\right)}{v \left(x\right)} ^ 2$

In this case $u \left(x\right) = \sin x$ and $v \left(x\right) = 1 - x {e}^{x}$

$u ' \left(x\right) = \cos x$ but to find $v ' \left(x\right)$ we will also need to use the product rule, which is given by:

$\frac{d}{\mathrm{dx}} g \left(x\right) h \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

$v ' \left(x\right) = - {e}^{x} - x {e}^{x}$

Now we have everything we need, so plug in the functions to obtain

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

This is about as simplified as it's going to get. We can split it into two fractions which could be useful in some scenarios:

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