# Question #829b0

Feb 1, 2018

Given $\ln \left(5 x - 3 {y}^{2}\right)$, $\frac{d}{\mathrm{dx}} \left[\ln \left(5 x - 3 {y}^{2}\right)\right] = \frac{5 - 6 y}{5 x - 3 {y}^{2}}$

#### Explanation:

$\ln \left(5 x - 3 {y}^{2}\right)$
First, you derive the inside of the function ($5 x - 3 {y}^{2}$)
1. $5 - 6 y$
Second, you put this answer over the inside of the function.
2. $\frac{d}{\mathrm{dx}} \left[\ln \left(5 x - 3 {y}^{2}\right)\right] = \frac{5 - 6 y}{5 x - 3 {y}^{2}}$.

Feb 1, 2018

#### Explanation:

.

Let $u = 5 x - 3 {y}^{2}$

$\mathrm{du} = 5 \mathrm{dx} - 6 y \mathrm{dy}$

$d \left(\ln \left(5 x - 3 {y}^{2}\right)\right) = d \left(\ln u\right) = \frac{1}{u} \mathrm{du} = \frac{5 \mathrm{dx} - 6 y \mathrm{dy}}{5 x - 3 {y}^{2}}$

If we only differentiate with respect to $x$ we will have:

$\frac{\mathrm{du}}{\mathrm{dx}} = 5$

$\frac{d}{\mathrm{dx}} \left(\ln \left(5 x - 3 {y}^{2}\right)\right) = \frac{d}{\mathrm{dx}} \left(\ln u\right) = \frac{1}{u} \mathrm{du} = \frac{5}{5 x - 3 {y}^{2}}$