# How do you use implicit differentiation to find dy/dx given cos2xsec3y+1=0?

Jul 19, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \tan \left(2 x\right)}{3 \tan \left(3 y\right)}$

#### Explanation:

Remember the product rule:

$\frac{d}{\mathrm{dx}} f \left(x\right) g \left(x\right) = f \left(x\right) g ' \left(x\right) + g \left(x\right) f ' \left(x\right)$

In this case we differentiate like this:

$\frac{d}{\mathrm{dx}} \left(\cos \left(2 x\right) \sec \left(3 y\right) + 1\right) = \frac{d}{\mathrm{dx}} 0$

$\frac{d}{\mathrm{dx}} \cos \left(2 x\right) \sec \left(3 y\right) = 0$

Let's say that $\cos \left(2 x\right) = f \left(x\right)$ and $\sec \left(3 y\right) = g \left(x\right)$. This gives us:

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \cos \left(2 x\right) = - \sin \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x\right) = - 2 \sin \left(2 x\right)$

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} \sec \left(3 y\right) = \sec \left(3 y\right) \tan \left(3 y\right) \cdot \frac{d}{\mathrm{dx}} 3 y = 3 \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \sec \left(3 y\right) \tan \left(3 y\right)$

Therefore we can differentiate our equation like this:

$\frac{d}{\mathrm{dx}} \cos \left(2 x\right) \sec \left(3 y\right) = 0$

$\cos \left(2 x\right) \left(3 \frac{\mathrm{dy}}{\mathrm{dx}} \sec \left(3 y\right) \tan \left(3 y\right)\right) + \sec \left(3 y\right) \left(- 2 \sin \left(2 x\right)\right) = 0$

All that is left to do is solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ in terms of $x$ and $y$.

$3 \cos \left(2 x\right) \sec \left(3 y\right) \tan \left(3 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} - 2 \sec \left(3 y\right) \sin \left(2 x\right) = 0$

$3 \cos \left(2 x\right) \sec \left(3 y\right) \tan \left(3 y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 2 \sec \left(3 y\right) \sin \left(2 x\right)$

dy/dx = (2sec(3y)sin(2x))/(3cos(2x)sec(3y)tan(3y)

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\sec}}} \left(3 y\right) \sin \left(2 x\right)}{3 \textcolor{red}{\cancel{\textcolor{b l a c k}{\cos}}} \left(2 x\right) \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{\sec}}} \left(3 y\right) \tan \left(3 y\right)} \cdot \frac{\frac{1}{\cos} \left(2 x\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\frac{1}{\cos} \left(2 x\right)}}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 \tan \left(2 x\right)}{3 \tan \left(3 y\right)}$