# How do you find (dy)/(dx) given ln(xy)=cos(y^4)?

Jun 23, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- y}{x + 4 x {y}^{4} \sin \left({y}^{4}\right)}$

#### Explanation:

Differentiate as normal, but remember that differentiating any function of $y$ with respect to $x$ will cause the chain rule to be in effect, essentially creating a $\frac{\mathrm{dy}}{\mathrm{dx}}$ term.

First, we can simplify the natural logarithm function using $\log \left(a b\right) = \log a + \log b$, so we don't have to use the product rule on $x y$.

$\ln x + \ln y = \cos \left({y}^{4}\right)$

Then, differentiating:

$\frac{1}{x} + \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = - 4 {y}^{3} \sin \left({y}^{4}\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

Group the $\frac{\mathrm{dy}}{\mathrm{dx}}$ terms, since we want to solve for it, the derivative:

$\frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {y}^{3} \sin \left({y}^{4}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{1}{y} + 4 {y}^{3} \sin \left({y}^{4}\right)\right) = - \frac{1}{x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \frac{1}{x}}{\frac{1}{y} + 4 {y}^{3} \sin \left({y}^{4}\right)}$

Multiplying the fraction by $\frac{x y}{x y}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- y}{x + 4 x {y}^{4} \sin \left({y}^{4}\right)}$