# How do you differentiate ysinx=y?

May 23, 2015

You can use the product rule, like the usual. Just remember that the derivative of y with respect to x in $\frac{\mathrm{df}}{\mathrm{dx}}$ is $\frac{\mathrm{dy}}{\mathrm{dx}}$.

With $f \left(x\right) : y \sin x = y$,

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

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[1 - \sin x\right] = y \cos x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \cos x}{1 - \sin x}$

If you check Wolfram Alpha, you'll see this, just multiplied by $- 1$.