How do I compute dy/dx given sin(6x + 2y) = xy?

Apr 6, 2018

The implicit derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$ is $\frac{y - 6 \cos \left(6 x + 2 y\right)}{2 \cos \left(6 x + 2 y\right) - x}$.

Explanation:

Implicitly differentiate both sides:

$\frac{d}{\mathrm{dx}} \left[\sin \left(6 x + 2 y\right)\right] = \frac{d}{\mathrm{dx}} \left[x y\right]$

$\cos \left(6 x + 2 y\right) \cdot \frac{d}{\mathrm{dx}} \left[6 x + 2 y\right] = \frac{d}{\mathrm{dx}} \left[x y\right]$

$\cos \left(6 x + 2 y\right) \cdot \left(6 + 2 \frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{d}{\mathrm{dx}} \left[x y\right]$

$6 \cos \left(6 x + 2 y\right) + 2 \cos \left(6 x + 2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[x y\right]$

$6 \cos \left(6 x + 2 y\right) + 2 \cos \left(6 x + 2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left[x\right] \cdot y + x \cdot \frac{d}{\mathrm{dx}} \left[y\right]$

$6 \cos \left(6 x + 2 y\right) + 2 \cos \left(6 x + 2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = y + x \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$2 \cos \left(6 x + 2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = y - 6 \cos \left(6 x + 2 y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 \cos \left(6 x + 2 y\right) - x\right) = y - 6 \cos \left(6 x + 2 y\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y - 6 \cos \left(6 x + 2 y\right)}{2 \cos \left(6 x + 2 y\right) - x}$

That's the derivative. Hope this helped!