# What is the derivative of f(x)=e^(4x)*ln(1-x) ?

Aug 27, 2014

$y ' = {e}^{4 x} \left(4 \ln \left(1 - x\right) - \frac{1}{1 - x}\right)$

Explanation:

f(x)=e^(4x)⋅ln(1−x)

Suppose, $y = f \left(x\right) \cdot g \left(x\right)$

In general Product Rule is,

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

Assume, $f \left(x\right) = {e}^{4 x}$ and $g \left(x\right) = \ln \left(1 - x\right)$

differentiating these functions with respect to $x$, we get

$f ' \left(x\right) = 4 \cdot {e}^{4 x}$

and $g ' \left(x\right) = - \frac{1}{1 - x}$

Plugging these in product rule definition yields,

$y ' = {e}^{4 x} \left(- \frac{1}{1 - x}\right) + \left(4 \cdot {e}^{4 x}\right) \ln \left(1 - x\right)$

$y ' = {e}^{4 x} \left(4 \ln \left(1 - x\right) - \frac{1}{1 - x}\right)$