Need help diffrenting trig problems In(x^2(3x-1))?

Jan 17, 2018

$\frac{d}{\mathrm{dx}} \left[\ln \left(\left({x}^{2}\right) \left(3 x - 1\right)\right)\right] = \frac{9 {x}^{2} - 2 x}{3 {x}^{3} - {x}^{2}}$

Explanation:

Well there are no trig functions in this problem. Perhaps you meant to say logarithms? Anyways...

Apply the following technique:

$\frac{d}{\mathrm{dx}} \left[\ln \left(u\right)\right] = \frac{1}{u} \cdot u '$

Given: $\ln \left({x}^{2} \left(3 x - 1\right)\right)$

I would distribute the ${x}^{2}$ to $3 x - 1$ and rewrite this as

$\ln \left(3 {x}^{3} - {x}^{2}\right)$

So when differentiating, I'll let $u = 3 {x}^{3} - {x}^{2}$ so now we'll find the deivative

$\frac{d}{\mathrm{dx}} \left[\ln \left(u\right)\right] = \frac{1}{3 {x}^{3} - {x}^{2}} \cdot \textcolor{red}{u '}$

$\textcolor{red}{u ' = 9 {x}^{2} - 2 x}$

So

$\frac{d}{\mathrm{dx}} \left[\ln \left(u\right)\right] = \frac{1}{3 {x}^{3} - {x}^{2}} \cdot 9 {x}^{2} - 2 x = \frac{9 {x}^{2} - 2 x}{3 {x}^{3} - {x}^{2}}$