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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.
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Implicit Differentiation Lesson 6.4
Tangent to a Circle • Consider the graph of the equation shown. GeogebraExample 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
Each time an expression has a y in it, we use the chain rule Use product rule and chain rule Implicit Differentiation • Given 4xy – 6y2 = 10 • We differentiate with respect to x on both sides of the equation 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
Note, this cannot be an equation, only an expression Implicit Differentiation on the TI Calculator • We can declare a function which will do implicit differentiation: • Usage:
Assignment • Lesson 6.4 • Page 401 • Exercises 1 – 21 odd, 43