1. Implicit Differentiation
Section 3.5

2. So far functions have been described by expressing one variable explicitly in terms of another variable—
y = or y = x sin x
Some functions are defined implicitly by a relation between x and y such as
x2 + y2 = or x3 + y3 = 6xy

3. Example; Solve for y x2 + y2 = 25
In some cases it is possible to solve such an equation for y as an explicit function (or several functions) of x.
Example; Solve for y x2 + y2 = 25
y =
The two functions are
f (x) = and g (x) =

4. Sometimes it is not so easy to explicitly solve for a variable.
Example;
x3 + y3 = 6xy
Equation of a curve called the folium of Descartes
It implicitly defines y as several functions of x.
When defined implicitly, it means the equation
x3 + [f (x)3] = 6xf (x)
is true for all values of in the domain.

5. Don’t need to solve an equation for y in terms of x to differentiate.
Instead use the method of implicit differentiation.
This requires differentiating both sides of the equation with respect to x and then solving for y .

6. Example 1; Differentiate
(a) If x2 + y2 = 25, find
(b) Find an equation of the tangent to the circle
x2 + y2 = 25 at the point (3, 4).
(a) Differentiate both sides of the equation x2 + y2 = 25:
y is a function of x use the Chain Rule,

7. Example 1 – Solution (b) At the point (3, 4) we have x = 3 and y = 4,
An equation of the tangent to the circle at (3, 4) is
y – 4 = (x – 3)
3x + 4y = 25

8. Example 2; Differentiate

9. Example 2; Find tangent line at point
tangent at point (3,3)
Tangent line horizontal

10. Example 2; Find tangent line at point
Tangent line horizontal

11. Example 3; Differentiate

12. Example 4; 2nd Derivative

13. Example 4; 2nd Derivative

14. Example 5; Differentiate

15. Example 6; Differentiate

16. 3.5 Implicit Differentiation
Summarize Notes
Read section 3.5
Homework
Pg.215 #1,9,11,17,23,25,27,35,49,51,53,76