Math 131. Implicit Differentiation Larson Section 2.5

Transcription

1 Math 131. Implicit Differentiation Larson Section.5 So far we have ealt with ifferentiating explicitly efine functions, that is, we are given the expression efining the function, such as f(x) = 5 x. However, equations can efine y as a function of x, for example the equation x + y = 9 leas to functions y = 5 x or y = 5 x. Sometimes equation may efine y as a function of x, but it may not be easy or even possible to solve for y. Nevertheless, when y is ifferentiable, we can fin the erivative for y without solving the equation for y by using implicit ifferentiation, which will involve ifferentiating the equation with respect to x, an then solving for. Before going into examples of implicit ifferentiation, let s remin ourselves of a few basics. Remember, if y is a ifferentiable function of x, then [y] =, while [x] = 1. Example 1. Suppose y is an unknown ifferentiable function of x. Evaluate the following erivatives: (a) [y3 ] () [sin(y)] (b) [x3 ] () [sin x] (e) (c) [x3 y 3 ] [x3 sin(y)] Solution: (a) This is just a generalize power rule [y3 ] = 3y [just think about what you woul o if you knew, for example, y = x + 1] (b) This is just the power rule, [x3 ] = 3x. (c) Now you have a prouct of two functions that you cannot simplify further, so it requires a prouct rule along with the answers from (a) an (b): [x3 y 3 ] = [x3 ] y 3 + x 3 [y3 ] = 3x y 3 + x 3 3y = 3x y 3 + 3x 3 y () This involves the chain rule since y is a function of x, so [sin y] = (cos y) (e) This is just the basic rule [sin x] = cos x (f) This uses the prouct rule along with erivatives alrea one: [x3 sin(y)] = [x3 ] sin(y) + x 3 [sin(y)] = 3x sin y + x 3 cos(y)

2 We will use techniques as in the previous example to ifferentiate both sies of an equation involving x an y with respect to x. Then we will solve the ifferentiate equation for. This process is calle implicit ifferentiation. The expression for will typically involve both x an y, to reuce it entirely to an equation in x woul require knowing what the function y is, an the point of implicit ifferentiation is to avoi neeing to know the expression for the function y. Example. (a) Differentiate the equation x + y = 5 implicitly to fin (b) Fin the slope of the tangent line to this circle at the point (3, 4). (c) Confirm your answer to (b), by solving the equation x + y = 5 for y, an then taking its erivative. Solution: (a) First ifferentiate both sies of the equation with respect to x: [x + y ] = [5] Now solve the last equation for : y x + y = 0 = x = x y = x y (b) We know = x, an at the point (3, 4), x = 3 an y = 4 so y = 3 4 of the tangent line. is the slope (c) Solving x + y = 5 for y we fin y = 5 x an so y = ± 5 x. At the point (3, 4), y > 0 (it is on the top half of the circle), so y = 5 x accoring to the chain rule = 1 (5 x ) 1/ x ( x) = 5 x when x = 3, we get = 3 = 3 is the slope of the tangent line. Notice that this agrees with (b), in fact we can check the erivative agrees with (b) because y = 5 x an so = x = x 5 x y A graph of the circle an tangent line is as follows

3 y x 4 In the previous example, it was easy to solve for y an so implicit ifferentiation coul have been avoie. However, that is not the case in the next few examples. Example 3. Fin the slope of the tangent line to the Lemniscate 3(x + y ) = 100(x y ) at the point (4, ). Solution: Differenting the equation implicitly yiels: an iviing both sies by 4 Then rearranging this becomes an solving for y yiels When x = 4 an y =, this means (x + y )(x + yy ) = 100(x yy ) 3(x + y )(x + yy ) = 50(x yy ) [3(x + y )y + 50y]y = 50x 3x(x + y ) y = Then the tangent line has slope /11. y = 50x 3x(x + y ) 3y(x + y ) + 50y 00 1(0) (0) = 40 0 = 11.

4 A very similar example to the previous is Example 4. the point (3, 1). Fin the slope of the tangent line to the curve (x + y ) = 5(x y ) at Solution: First, ifferentiate implicitly an so (x4 + ()x y + y 4 ) = (5x 5y ) 4()x 3 + 4()xy + 4()x yy + 4()y 3 y = (5)x (5)yy Diviing both sies by, an rearranging yiels (4x y + 4y 3 + 5y)y = 5x 4x 3 4xy an so = 5x 4x3 4xy 4x y + 4y 3 + 5y. When x = 3 an y = 1 we have = 45 5 Example 5. Fin when 3x5 + 9x 9 y 3y + sin(y) = 1. Solution: Differentiating with respect to x we fin 15x x 8 y + 9x 9 y 18y 5 y + y cos y = 0. Next we wish to isolate y on one sie of the equation Now solve for y to fin = (9x 9 18y 5 + cos y)y = 15x 4 81x 8 y 15x4 81x 8 y 9x 9 18y 5 + cos y = 15x4 + 81x 8 y 18y 5 9x 9 cos y The next few examples involve easier equations, but ask for tangent lines or where certain tangent lines are on the graph. Example. Fin y when y is implicitly efine by 4x + y + 3xy = 10 +

5 Then fin the slope of the tangent line to this curve at the point (, 1). Solution: Differentiating implicitly we have 1 (4x + y) 1/ (4 + y ) + 1 (3xy) 1/ (3y + 3xy ) = 0 an so ( + 3x ) ( y = 4x + y 3xy 4 + 4x + y 3y ) 3xy then y = 4 4x+y + 3y 3xy 4x+y + 3x 3xy Plugging in x = an y = 1 we have the slope of the tangent line at (, 1) is y = Example 7. Fin the equation of the tangent line to the graph at the given point. (x + 3) + (y ) =, ( 8, 3) Solution: Using implicit ifferentiation, we fin [(x + 3) + (y ) ] = [] (x + 3) + (y ) = 0 an therefore, (y ) = (x + 3) an so At the point ( 8, 3), x = 8 an y = 3 an so = = 5 = (x + 3) (y ). The the tangent line has equation y (3) = (5)(x ( 8)) which in slope-intercept form is y = 5x + 43 A graph of the circle an tangent line is as follows

6 8 y 4 x Example 8. Fin all points at which the graph of the following equation has a vertical or horizontal tangent line. 9x + 4y 3x + 40y = 100 Solution: Using implicit ifferentiation, we fin [9x + 4y 3x + 40y] = [ 100] 18x + 8y = 0 solving the latter equation for implies [8y + 40] The horizontal tangent lines occur when 18x + 3 8y + 40 = 18x + 3 8y = 0, an so = 18x + 3 an so = 0 18x + 3 = 0 x = 3 18 = When x =, this implies 9() + 4y 3() + 40y = 100 an so 4y + 40y + 4 = 0 This quaratic equation has solutions y = 8 an y =. The points where the tangent lines are horizontal are then (, 8) an (, ). The graph of the equation is an ellipse, therefore it has vertical tangent lines where the erivative is unefine, thus when 8y + 40 = 0, or y = 40/8 = 5. We plug y = 5 back into the original equation to solve for x: 9x + 4( 5) 3x + 40( 5) = 100 9x 3x = 0

7 This quaratic equation has solutions x = 0 an x = 4. This means that the graph has vertical tangent lines at the points (0, 5) an (4, 5). Precalculus note. This problem coul have been solve using precalculus: completing the squares you can write the equation of the ellipse in stanar form an then graph it. In this case, the major axis is vertical, an so the horizontal tangent lines are at the en points of the major axis an the vertical tangent lines are at the en points of the minor axis. Example 9. Fin the equations of the two lines tangent to the ellipse x 1 + y 9 = 1 that pass through the point (8, 0). Solution: To fin the equations of the two lines tangent to the ellipse x 1 + y 9 = 1 that pass through the point (8, 0), we first ifferentiate x 1 + y 9 = 1 implicitly to fin y : x 1 + yy 9 = 0 an so y = x 1 9 y = 9x 1y A line tangent to the ellipse at the point (x 0, y 0 ) passing through the point (8, 0) must have slope m = y 0 0 x 0 8 = 9x 0 1y 0 This implies 1y 0 = 9x 0 + (8)(9)x 0 an so (1)(9) 9x 0 = 9x 0 + (8)(9)x 0 Thus 1 = 8x 0 an so x 0 = 1 8 = an then y 0 = ± 3 3. Thus, the tangent lines pass ( ) ( ) through the points, 3 3 an, 3 3, an have slopes 3 3 an 3 3 respectively. (3)(4) (3)(4) The equations of the tangent lines (which the reaer can rewrite or simplify as esire) will be x 3y 8 + x 3y = 1 an 8 = 1