Implicit Differentiation and Inverse Trigonometric Functions

Transcription

1 Implicit Differentiation an Inverse Trigonometric Functions MATH 161 Calculus I J. Robert Buchanan Department of Mathematics Summer 2018

2 Explicit vs. Implicit Functions y 0.0 y y = x 2 + 2x 3 x 2 + y 2 = 1

3 Motivation (1 of 2) Think of y as a function y(x) then x 2 + y 2 = 1 x 2 + (y(x)) 2 = 1 (y(x)) 2 = 1 x 2 y(x) = 1 x 2 or y(x) = 1 x 2. The one equation efines two implicit functions of x.

4 Motivation (2 of 2) if y 0. [ x 2 + y 2] = x x [1] [ x 2 + (y(x)) 2] = 0 x 2x + 2y(x) y x 2y y x y x = 0 = 2x = x y

5 Implicit Differentiation The process of ifferentiating both sies of an equation is known as implicit ifferentiation. When we encounter a function of y, where y is implicitly a function of x, we use the following erivative formula (the Chain Rule): x [g(y)] = g (y) y x

6 Example Fin y x if x ey 3y sin x = 1.

7 Example Fin y x if x ey 3y sin x = 1. x [x ey 3y sin x] = x [1]

8 Example Fin y x if x ey 3y sin x = 1. x [x ey 3y sin x] = x [1] e y + x e y y 3y sin x 3y cos x = 0

9 Example Fin y x if x ey 3y sin x = 1. x [x ey 3y sin x] = x [1] e y + x e y y 3y sin x 3y cos x = 0 (xe y 3 sin x)y = 3y cos x e y y x = 3y cos x ey xe y 3 sin x

10 Example Fin the slope of the tangent line to the graph of x 3 y 2 4 x = x 2 y at (x, y) = (4, 1/2) y x

11 Solution x 3 y 2 4 x = x 2 y [ x 3 y 2 4x 1/2] = [ ] x 2 y x x

12 Solution x 3 y 2 4 x = x 2 y [ x 3 y 2 4x 1/2] = x x [ x 2 y 3x 2 y 2 + 2x 3 y y 2 x = 2xy + x 2 y ]

13 Solution x 3 y 2 4 x = x 2 y [ x 3 y 2 4x 1/2] = x x [ x 2 y 3x 2 y 2 + 2x 3 y y 2 x = 2xy + x 2 y y 1 = y when x = 4, y = 1/2 ]

14 Solution x 3 y 2 4 x = x 2 y [ x 3 y 2 4x 1/2] = x x [ x 2 y 3x 2 y 2 + 2x 3 y y 2 x = 2xy + x 2 y y 1 = y when x = 4, y = 1/2 y = 7 48 ]

15 Example Fin the equation of the tangent line to the graph of x 3 y + y 3 x 1.0 = 1 at (x, y) = (1/2, ) y x

16 Solution (1 of 2) x x 3 y + y 3 x [ x 3 y + y 3 ] x = 1 = x [1]

17 Solution (1 of 2) x x 3 y + y 3 x [ x 3 y + y 3 ] x = 1 = x [1] 3x 2 y x 3 y y 2 + 3y 2 y x y 3 x 2 = 0

18 Solution (1 of 2) x x 3 y + y 3 x [ x 3 y + y 3 ] x = 1 = x [1] 3x 2 y x 3 y y 2 + 3y 2 y x y 3 x 2 = 0 when (x, y) = (1/2, ). y =

19 Solution (2 of 2) m = y y 0 x x = y x 1/2 y = x

20 Example Fin the horizontal an vertical tangents to the graph of the equation xy 2 2y = 2.

21 Example Fin the horizontal an vertical tangents to the graph of the equation xy 2 2y = 2. First we must fin y : [ ] xy 2 2y x = x [2] y 2 + 2xy y 2y = 0 y 2 + 2(xy 1)y = 0 y = y 2 2(1 xy)

22 Tangents 2 = xy 2 2y y = y 2 2(1 xy) Tangent lines are horizontal when y = 0 which implies y = 0. However, when y = 0 the first equation cannot be satisfie. Thus there are no points on the curve where the tangent line is horizontal.

23 Tangents 2 = xy 2 2y y = y 2 2(1 xy) Tangent lines are horizontal when y = 0 which implies y = 0. However, when y = 0 the first equation cannot be satisfie. Thus there are no points on the curve where the tangent line is horizontal. Tangent lines are vertical when y is unefine. This implies xy = 1 or equivalently x = 1/y. Substituting this into the first equation yiels 2 = y 2y = y = (x, y) = ( 1/2, 2).

24 Illustration 2 = xy 2 2y y x

25 Example Fin y given that x 3 + y 3 = 1.

26 Example Fin y given that x 3 + y 3 = 1. First we must fin y. [ x 3 + y 3] x = x [1] 3x 2 + 3y 2 y = 0 y = x 2 y 2

27 Example Fin y given that x 3 + y 3 = 1. First we must fin y. [ x 3 + y 3] x = x [1] 3x 2 + 3y 2 y = 0 y = x 2 Then we must ifferentiate a secon time. y 2

28 Secon Derivative So far we have: x 3 + y 3 = 1 y = x 2 When we ifferentiate again we fin, [ y ] = [ x 2 ] x x y 2 y 2

29 Secon Derivative So far we have: x 3 + y 3 = 1 y = x 2 When we ifferentiate again we fin, [ y ] = [ x 2 ] x x y 2 y 2 y = 2xy 2 x 2 (2y)y (y 2 ) 2 (quotient rule)

30 Secon Derivative So far we have: x 3 + y 3 = 1 y = x 2 When we ifferentiate again we fin, [ y ] = [ x 2 ] x x y 2 y 2 y = 2xy 2 x 2 (2y)y (y 2 ) 2 (quotient rule) = 2xy 2 + 2x 2 y( x 2 /y 2 ) y 4 = 2xy 2 2x 4 /y y 4 = 2xy 3 2x 4 y 5

31 Right Triangle Trigonometry Suppose θ = sin 1 x or equivalently sin θ = x, then we can picture θ as an acute angle in a right triangle. 1 x θ=sin -1 x 1 - x 2

32 Derivative of f (x) = sin 1 x We can use the Chain Rule to fin erivatives of the inverse trigonometric functions. sin(sin 1 x) = x [ ] sin(sin 1 x) = x x [x] cos(sin 1 x) [ ] sin 1 x = 1 x 1 x 2 [ ] sin 1 x = 1 x [ ] sin 1 1 x = x 1 x 2

33 Derivatives of the Inverse Trigonometric Functions x x x x x x [ ] sin 1 x [ ] cos 1 x [ ] tan 1 x [ ] sec 1 x [ ] cot 1 x [ ] csc 1 x = 1 1 x 2 1 = 1 x 2 1 = 1 + x 2 1 = x x 2 1 = x 2 1 = x x 2 1

34 Examples Fin the following erivatives: [ ] 1. cos 1 3x x 2. [sin 1 e x] x 3. [tan 1 x 2] x [ ] 4. sec 1 ln x x

35 Examples Fin the following erivatives: [ ] 1. cos 1 3 3x = x 1 9x 2 2. [sin 1 e x] x 3. [tan 1 x 2] x [ ] 4. sec 1 ln x x

36 Examples Fin the following erivatives: [ ] 1. cos 1 3 3x = x 1 9x 2 [ 2. sin 1 e x] e x = x 1 e 2x 3. [tan 1 x 2] x [ ] 4. sec 1 ln x x

37 Examples Fin the following erivatives: [ ] 1. cos 1 3 3x = x 1 9x 2 [ 2. sin 1 e x] e x = x 1 e 2x [ 3. tan 1 x 2] = 2x x 1 + x 4 4. x [ ] sec 1 ln x

38 Examples Fin the following erivatives: [ ] 1. cos 1 3 3x = x 1 9x 2 [ 2. sin 1 e x] e x = x 1 e 2x [ 3. tan 1 x 2] = 2x x 1 + x 4 [ ] 4. sec 1 1/x ln x = x ln x (ln x) 2 1

39 Homework Rea Section 2.8 Exercises: 1 27 o (implicit ifferentiation), o (inverse trigonometric functions)