How do you use implicit differentiation to find dy/dx given xy+y=sinx?

Dec 2, 2016

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

Explanation:

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

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

Now differentiate using the product rule on the first term.

$y + x \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} = \cos x$

Then rearrange for $\text{ } \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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