# How do you find the derivatives of z=ln(ysqrt(3y+1))?

Jun 16, 2017

$\frac{\mathrm{dz}}{\mathrm{dy}} = \frac{9 y + 2}{6 {y}^{2} + 2 y}$

#### Explanation:

We have:

$z = \ln \left(y \sqrt{3 y + 1}\right)$

Using the rules of logs:

$\log a b = \log a + \log b$, and, $\log {a}^{b} = b \log a$

we can rewrite the expression as:

$z = \ln y + \ln \sqrt{3 y + 1}$
$\setminus \setminus = \ln y + \ln {\left(3 y + 1\right)}^{\frac{1}{2}}$
$\setminus \setminus = \ln y + \frac{1}{2} \setminus \ln \left(3 y + 1\right)$

Then using the standard Calculus result:

$\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$

along with the chain rule, we have:

$\frac{\mathrm{dz}}{\mathrm{dy}} = \frac{1}{y} + \frac{1}{2} \cdot \frac{1}{3 y + 1} \cdot 3$
$\text{ } = \frac{1}{y} + \frac{3}{2} \cdot \frac{1}{3 y + 1}$

We could if required simplify this further:

$\frac{\mathrm{dz}}{\mathrm{dy}} = \frac{2 \left(3 y + 1\right) + 3 y}{2 y \left(3 y + 1\right)}$
$\text{ } = \frac{6 y + 2 + 3 y}{6 {y}^{2} + 2 y}$
$\text{ } = \frac{9 y + 2}{6 {y}^{2} + 2 y}$