1. Implicit Differentiation
Lesson 6.4

2. Tangent to a Circle Consider the graph of the equation shown.
Geogebra Example
How can we use calculus to find the slope of a tangent for a particular (x, y) on the circle?
Why is this a problem?

3. Explicit Functions We have worked with functions of the form Examples:
Even when given 2x – 3y = 12
We can solve for

4. Implicit Functions Some functions cannot be readily solved for y.
For these we say y is given implicitly in terms of x
We will use implicit differentiation to find

5. Implicit Differentiation
Given 4xy – 6y2 = 10
We differentiate with respect to x on both sides of the equation
Each time an expression has a y in it, we use the chain rule
Use product rule and chain rule
Use chain rule

6. Implicit Differentiation
Now we have an equation and solve for

7. We Better Try This Again
Find dy/dx for following

8. Tangent Lines Consider the equation for a circle
x2 + y2 = 36
What is the equation of the tangent to the circle at the point where x = 5 in the 4th quadrant
Find the slope by using implicit differentiation
Substitute in (5, )
Use point-slope formula for line

9. Review To find dy/dx for an equation containing both x and y
Differentiate both sides of equation w/respect to x
Assuming y is a function of x
Place all terms with dy/dx on one side
All others on other side
Factor out dy/dx
Solve for dy/dx

10. Implicit Differentiation on the TI Calculator
We can declare a function which will do implicit differentiation:
Usage:
Note, this cannot be an equation, only an expression

11. Assignment
Lesson 6.4
Page 401
Exercises 1 – 21 odd, 43