# How do you find dy/dx by implicit differentiation of x=sec(1/y)?

Dec 10, 2017

Given:

$x = \sec \left(\frac{1}{y}\right)$

Differentiate both sides:

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

The left side is 1:

$1 = \frac{d \left(\sec \left(\frac{1}{y}\right)\right)}{\mathrm{dx}}$

The right side requires the recursive use of the chain rule:

let $u = \frac{1}{y}$ then

$\frac{d \left(\sec \left(\frac{1}{y}\right)\right)}{\mathrm{dx}} = \frac{\mathrm{ds} e c \left(u\right)}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{ds} e c \left(u\right)}{\mathrm{du}} = \tan \left(u\right) \sec \left(u\right)$

Reverse the substitution:

$\frac{\mathrm{ds} e c \left(u\right)}{\mathrm{du}} = \tan \left(\frac{1}{y}\right) \sec \left(\frac{1}{y}\right)$

$\frac{\mathrm{du}}{\mathrm{dy}} = - \frac{1}{y} ^ 2$

Returning to the equation:

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

Solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$ in 3 steps:

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

$- {y}^{2} \cot \left(\frac{1}{y}\right) = \sec \left(\frac{1}{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{2} \cot \left(\frac{1}{y}\right) \cos \left(\frac{1}{y}\right)$