# How do you differentiate sin(xy)=ln(x/y)?

Oct 10, 2016

The question does not specify the derivative with respect to what? Here are two answers.

#### Explanation:

$\sin \left(x y\right) = \ln \left(\frac{x}{y}\right)$. First, let's rewrite that $\ln$.

$\sin \left(x y\right) = \ln \left(x\right) - \ln \left(y\right)$

Here is the general case first:

The derivative with respect to $t$

$\frac{d}{\mathrm{dt}} \left(\sin \left(x y\right)\right) = \frac{d}{\mathrm{dt}} \left(\ln \left(x\right)\right) - \frac{d}{\mathrm{dt}} \left(\ln \left(y\right)\right)$

Evaluating the derivatives using the chain rule, we get,

$\cos \left(x y\right) \frac{d}{\mathrm{dt}} \left(x y\right) = \frac{1}{x} \frac{d}{\mathrm{dt}} \left(x\right) - \frac{1}{y} \frac{d}{\mathrm{dt}} \left(y\right)$.

$\cos \left(x y\right) \left[\frac{\mathrm{dx}}{\mathrm{dt}} y + x \frac{\mathrm{dy}}{\mathrm{dt}}\right] = \frac{1}{x} \frac{\mathrm{dx}}{\mathrm{dt}} - \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dt}}$.

Solve algebraically for whichever one you're interested in.

If you want to find the derivative of $y$ with respect to $x$, then you're looking for $\frac{\mathrm{dy}}{\mathrm{dx}}$. You could replace the $t$'s above with $x$'s, but it's probably more clear if we just start over.

The derivative with respect to $x$

$\frac{d}{\mathrm{dx}} \left(\sin \left(x y\right)\right) = \frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) - \frac{d}{\mathrm{dx}} \left(\ln \left(y\right)\right)$

Evaluating the derivatives using the chain rule, we get,

$\cos \left(x y\right) \frac{d}{\mathrm{dx}} \left(x y\right) = \frac{1}{x} - \frac{1}{y} \frac{d}{\mathrm{dx}} \left(y\right)$.

$\cos \left(x y\right) \left[y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right] = \frac{1}{x} - \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}}$.

Before solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$ it is helpful to get rid of the fractions by multiplying both sides of the equation by $x y$.

$x y \cos \left(x y\right) \left[y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right] = y - x \frac{\mathrm{dy}}{\mathrm{dx}}$.

$x {y}^{2} \cos \left(x y\right) + {x}^{2} y \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = y - x \frac{\mathrm{dy}}{\mathrm{dx}}$.

$x \frac{\mathrm{dy}}{\mathrm{dx}} + {x}^{2} y \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = y - x {y}^{2} \cos \left(x y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - x {y}^{2} \cos \left(x y\right)}{x + {x}^{2} y \cos \left(x y\right)}$