Home > APCALC > Chapter 6 > Lesson 6.2.2 > Problem6-71

6-71.

Given the equation $x^2 - 2xy + \tan(y) = 4$, what is $\frac { d y } { d x }$?

To avoid the preliminary step of solving for $y$, you could Implicitly differentiate:

$\frac{d}{dx}(x^{2}-2xy+\text{tan}y)=\frac{d}{dx}4$

$\frac{dx}{dx}2x-\frac{dx}{dx}2y-\frac{dy}{dx}2x+\frac{dy}{dx}\text{sec}^{2}y=0$

$\text{Solve for }\frac{dy}{dx}.$