AP Calculus BC / Unit 3: Differentiation: Composite, Implicit, and Inverse Functions
3.2 Implicit Differentiation
Notes
When $y$ isn't solved for, differentiate both sides with respect to $x$, treating $y$ as a function of $x$: every $y$ term gets a factor of $\displaystyle \frac{dy}{dx}$. Then solve for $\displaystyle \frac{dy}{dx}$.
- $\displaystyle \frac{d}{dx}y^2=2y\frac{dy}{dx}$
- $\displaystyle \frac{d}{dx}(xy)=y+x\frac{dy}{dx}$
Practice
For $x^2+y^2=25$, $\displaystyle \frac{dy}{dx}=$
$2x+2yy'=0$.
Using $x^2+y^2=25$, the slope at $(3,4)$ is:
$\displaystyle -\frac{3}{4}$.
$\displaystyle \frac{d}{dx}(xy)=$
Product rule, with $y'$ for the derivative of $y$.
Ready for more? Take the Unit 3 test.