Question

Find the derivative of the function

f(x)=(2x+1)ln(2x+7)
and then evaluate it at x=0 .

f′(0)=

Answer 1

$f'(x)=2\ln(2x+7) + \left(\dfrac{2x+1}{2x+7}\right)$

$f'(0) = 2\ln{7} + \frac{1}{7}$

Step-by-step explanation:

Recall that the derivative of a product of two functions $f(x)=g(x)h(x)$ is defined as

$\dfrac{df(x)}{dx}=f'(x)= \dfrac{dg(x)}{dx}h(x) + g(x)\dfrac{dh(x)}{dx}$

Let $g(x)=2x+1$ and $h(x)=\ln(2x+7)$ so f'(x) is

$f'(x)=2\ln(2x+7) + \left(\dfrac{2x+1}{2x+7}\right)$

Evaluated at x = 0, the value of the derivative is

$f'(0) = 2\ln{7} + \frac{1}{7}$