2.7 Implicit Differentiation

Transcription

1 2.7 Implicit Differentiation [ y] = = = [ x] 1 Sometimes we may be intereste in fining the erivative of an equation that is not solve or able to be solve for a particular epenent variable explicitly. In this case, we must evelop a way to analyze that variable s rate of change implicitly. The equations below are equivalent, but are in ifferent forms with respect to the epenent variable y. Example 1: Explicit form Implicit form 2 y = x xy = 2 Fin the erivative,, of 2 y =, an equation explicitly solve for y by taking the erivative of both sies x with respect to x. Example 2: Fin the erivative,, of xy = 2, an equation NOT explicitly solve for y by taking the erivative of both sies with respect to x. Solve for. The metho use to fin in Example 2 is calle Implicit Differentiation. What o you notice about the implicit result compare to the result from Example 1? Which metho was easier/faster? Page 1 of 6

2 In the first two examples, we ha the option of ifferentiating explicitly or implicitly, but most of the time, we will use implicit ifferentiation when we re ealing with equations of curves that are not functions of a single variable, whose equations have powers of y greater than 1 making it ifficult or impossible to explicitly solve for y. For such equations, we will be force to use implicit ifferentiation, then solve for, which will be a function of either y alone or both x an y. Important note 1: Just because an equation is not explicitly solve for a epenent variable oesn t mean it can t. Your first step is to analyze whether it can be solve explicitly. This will make the problem a bit easier an your erivative will be a function of a single variable. Important note 1.5: When ifferentiation implicitly, you must show that you are taking the erivative of both sies with respect to x. [ Left Sie] = [ Right Sie] Important note 2: It matters not if you use or y, just be careful not to mistake y for 1 y = y. stans out more prominently, but y is a wee ta little bit faster. Example 3: For the equation y x sec( y) = +, fin. In Example 3, we were aske to fin the variable equation for. Sometimes, however, we are intereste only in fining the slope of a curve at given point. For such a problem, we can save a bit of algebraic manipulation if we plug in (an inicate so) our point earlier. Example 4: Fin the slope of the graph of ( 1, 3) 3 + = at ( 1, 3) y y 5y x 4 by fining first, then evaluating at Page 2 of 6

3 Example 5: Fin the slope of the graph of before solving for. 3 + = at ( 1, 3) by ifferentiating then plugging in ( 1, 3) y y 5y x 4 Notice that in Examples 4 an 5, because our erivative equation ha both an x an a y in it, we neee the actual orere pair ( xy, ) to evaluate. Often when only an x-value is given, an not an orere pair, it is sign that our given equation can be explicitly solve for a single equation of y, like in Examples 1 an 2, but not always. Example 6: Fin at x = 1 for the equation 2 y + x= 2xy When your erivative is a quotient of two variable expressions, then 0 Horizontal tangents occur when = 0 0 Vertical tangents occur when = 0 0 No tangent line exists when = 0 (these points must be thrown out) Page 3 of 6

4 Example 7: The graph of the equation x + y = 4 is a circle centere at the origin with a raius of two. (a) Sketch the graph of the equation (a) fin. (b) Using calculus, verify that the graph has horizontal tangents at the point ( 0, 2 ) an ( 0, 2). (c) Using calculus, verify that the graph has vertical tangents at the point ( ) 2,0 an( 2,0). () At what value(s) of x is the slope of the graph 3 4? Page 4 of 6

5 Example 8: Graph generate by WolframAlpha The equation for the graph of the rotate ellipse shown above is 2x xy 4y =, (a) Determine the x-value(s) of any horizontal tangent lines to the graph of 2x xy 4y =. (b) Determine the y-value(s) of any vertical tangent lines to the graph of 2x xy 4y =. Page 5 of 6

6 Example 9: Determine the equation of the tangent line of ( ) 2 + = at the point ( 3,1 ) 3 x y 100xy When aske to fin a higher-orer erivative where implicit ifferentiation is neee, it is always beneficial to solve for prior to fining the secon erivative an beyon. This will always be possible because the first erivative will be a linear function of. Subsequent erivatives will have a visible in the equation, at which time, a substitution may be mae to put your final erivative in terms of x an y. Example 10: 2 y Fin as a function of x an y if x 3y = 8. Page 6 of 6