Implicit Differentiation Section 3.5

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 = 25 or x3 + y3 = 6xy

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) = .

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.

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 .

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,

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

Example 2; Differentiate

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

Example 2; Find tangent line at point Tangent line horizontal

Example 3; Differentiate

Example 4; 2nd Derivative

Example 4; 2nd Derivative

Example 5; Differentiate

Example 6; Differentiate

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