# Thread: Implicit Differentiation of Trig functions

1. ## Implicit Differentiation of Trig functions

How would one go about doing these? I know how to do implicit differentiation, just not of trig functions. Specifically I'm trying to find y sec(x) = x tan(y)

2. Hello, fattydq!

I must assume you know the trig derivatives . . . and the Product Rule.

$y\cdot\sec(x) \:=\:x\cdot\tan(y)$

. . . . . . $\underbrace{y\cdot\sec(x)}_{\text{Product Rule}} \qquad\qquad = \qquad\quad\;\; \underbrace{x\cdot\tan(y)}_{\text{Product rule}}$

$\overbrace{y\cdot\sec(x)\tan(x) + y'\cdot\sec(x)} \;=\; \overbrace{x\cdot\sec^2(y)\cdot y' + 1\cdot\tan(y)}$

. . $\sec(x)\cdot y' - x\sec^2(y)\cdot y' \;=\;\tan (y) - y\sec(x)\tan(x)$

Factor: . $\bigg[\sec(x) - x\sec^2(y)\bigg]\cdot y' \;=\;\tan(y) - y\sec(x)\tan(x)$

Therefore: . $y' \;=\;\frac{\tan(y) - y\sec(x)\tan(x)}{\sec(x) - x\sec^2(y)}$