# How do you differentiate 1/cos(x) = x/y^2-y?

Jul 17, 2016

Use implicit differentiation and algebra to get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left(1 - \sec \left(x\right) \tan \left(x\right) {y}^{2}\right)}{2 x + {y}^{3}}$

#### Explanation:

Assume that the given equation, which is equivalent to $\sec \left(x\right) = \frac{x}{y} ^ 2 - y$, implicitly defines $y$ as a function of $x$ (the assumption is true wherever the graph of this equation does not have a vertical tangent line).

Now differentiate both sides with respect to $x$, keeping in mind the assumption we made while also using the quotient rule and chain rule:

$\sec \left(x\right) \tan \left(x\right) = \frac{{y}^{2} - 2 x y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 4 - \frac{\mathrm{dy}}{\mathrm{dx}}$.

This simplifies to

$\sec \left(x\right) \tan \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \left(- \frac{2 x + {y}^{3}}{y} ^ 3\right) + \frac{1}{y} ^ 2$

Now solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ to get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\sec \left(x\right) \tan \left(x\right) - \frac{1}{y} ^ 2\right) \cdot \left(- {y}^{3} / \left(2 x + {y}^{3}\right)\right)$

$= \frac{1 - \sec \left(x\right) \tan \left(x\right) {y}^{2}}{y} ^ 2 \cdot {y}^{3} / \left(2 x + {y}^{3}\right)$

$= \frac{y \left(1 - \sec \left(x\right) \tan \left(x\right) {y}^{2}\right)}{2 x + {y}^{3}}$