Implicit Differentiation Lesson 6.4

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?

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

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

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

Implicit Differentiation Now we have an equation and solve for

We Better Try This Again Find dy/dx for following

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, -3.316) Use point-slope formula for line

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

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

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