# Thread: derivative with natural logs.

1. ## derivative with natural logs.

just wanted to check to see if i'm doing this right, any step by step guidance is greatly appreciated.

"find the derivative":
$
f(x) = ln(x^2 + 3x + 1) ln(x^3+1)$

would i start by using chain rule?

u= $x^2 + 3x + 1$

= $2x +3$

$h(u) = lnu = \frac{1}{u}$

$= \frac{1}{x^2+3x+1} * 2x+3 = \frac{2x+3}{x^2+3x+1}$

u= $x^3 + 1 = 3x^2$

$h(u) = lnu = \frac{1}{u}$

$= \frac{1}{x^3+1} * 3x^2 = \frac{3x^2}{x^3+1}$

since those are fractions would i use the quotient rule? or just plug them the way they are into product rule?

with product rule i have:

$f\prime(x) = \frac{2x+3}{x^2+3x+1} * ln(x^3+1) + ln(x^2+3x+1) * \frac{3x^2}{x^3+1}
$

help please! im so confused, final exam tomorrow =\

2. Your method is completely correct.
You're given $f(x) = uv$, hence $f'(x)=u'v+v'u$ and you have found u' and v' using the chain rule.

3. Use the product rule here

for $f=u\times v \implies f' = uv'+vu'$

Also $(\ln f(x))' = \frac{f'(x)}{f(x)}$