What is the implicit derivative of 1=x/y-cosy?

Sep 2, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{{y}^{2} \sin y - x}$

Explanation:

By the quotient rule and implicit differentiation:

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

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

$0 = \frac{1 \times y - 1 \times x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{y} ^ 2 + \sin y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

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

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

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{- {y}^{2} \sin y + x}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{{y}^{2} \sin y - x}$

Hopefully this helps!