Implicit Differentiation Section 3 5 So far functions

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 x 2 + y 2 = 25 or x 3 + y 3 = 6 xy

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 x 2 + y 2 = 25 y= The two functions are f ( x) = and g (x ) = .

Sometimes it is not so easy to explicitly solve for a variable. Example; x 3 + y 3 = 6 xy 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 x 3 + [f (x)3] = 6 xf (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 x 2 + y 2 = 25, find . (b) Find an equation of the tangent to the circle x 2 + y 2 = 25 at the point (3, 4). (a) Differentiate both sides of the equation x 2 + y 2 = 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) 3 x + 4 y = 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; 2 nd Derivative

Example 4; 2 nd 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