# How do you find the derivatives of y=ln(x^2y)?

Aug 19, 2017

#### Explanation:

To find the derivatives of $x$ and $y$ with respect to some variable $t$, use the chain rule (implicit differentiation).

$y = \ln \left({x}^{2} y\right)$ I would rewrite using properties of logarithms.

$y = 2 \ln x + \ln y$

Differentiate w.r.t. $t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{x} \frac{\mathrm{dx}}{\mathrm{dt}} + \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dt}}$

Solving for $\frac{\mathrm{dy}}{\mathrm{dt}}$

$\left(1 - \frac{1}{y}\right) \frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{x} \frac{\mathrm{dx}}{\mathrm{dt}}$

$\frac{y - 1}{y} \frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{x} \frac{\mathrm{dx}}{\mathrm{dt}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2 y}{x \left(y - 1\right)} \frac{\mathrm{dx}}{\mathrm{dt}}$

And solving for $\frac{\mathrm{dx}}{\mathrm{dt}}$,

$\left(1 - \frac{1}{y}\right) \frac{\mathrm{dy}}{\mathrm{dt}} = \frac{2}{x} \frac{\mathrm{dx}}{\mathrm{dt}}$

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{x}{2} \left(1 - \frac{1}{y}\right) \frac{\mathrm{dy}}{\mathrm{dt}}$

$= \frac{x \left(y - 1\right)}{2 y} \frac{\mathrm{dy}}{\mathrm{dt}}$