2.8  IMPLICIT  FUNCTIONS

We now turn to the topic of implicit differentiation. We say that y is an implicit function of x if we are given an equation

Which determines y as a function of x. An example is x+ xy=2y. Implicit differentiation is a way of finding the derivative of y without actually solving for y as a function of x, assume that dy/dx exists. The method has two steps:

Step 1   Differentiate both sides of the equation σ (x, y) = ι (x,y) to get a new equation

(1)            

The Chain Rule is often used in this step.

Step 2  Solve the new Equation 1 for dy/dx. Then answer will usually involve both x and y.

       In each of the examples below, we assume that dy/dx exists and use implicit differentiation to find the value of dy/dx.

EXAMPLE  1  Given the equation  x+ xy =2y, find dy/dx.

Step 1

          We find each side by the Sum and Product Rules,

Thus our new equation is

Step 2  Solve for dy/dx.

We can check our answer by solving the original equation for y and using ordinary differentiation:

By the Quotient Rule,

A third way to find dy/dx is to solve the original equation for x, find dx/dy, and then use the Inverse Function Rule.

To see that our three answers

are all the same we substitute ___________ for y :

In Example 1, we found dy/dx by three different methods.

(a) Implicit differentiation. We get dy/dx in terms of both x and y.

(b) Solve for y as a function of x and differentiate directly. This gives dy/dx in terms of x only.

(c) Solve for x as a function of y, find dx/dy directly, and use the Inverse Function Rule. This method gives 

dy/dx in terms of y only.

EXAMPLE  2  Given y+ ____ = x², find dy/dx.

This answer can be used to find the slope at any point on the curve. For example, at the point (_____, 1) the slope is

While at the point (_______, 1) the slope is

To get dy/dx in terms of x, we solve the original equation for y using the quadratic formula:

Since _____ , only one solution may occur,

Then                       

The graph of this function is shown in Figure 2.8.1. By substitution we get

Figure 2.8.1

      Often one side of an implicit function equation is constant and has derivative zero.

EXAMPLE  3  Given x² - 2y² = 4, y ≤ 0, find dy/dx.

Solving the original equation for y, we get

Thus dy/dx in terms of x is

The graph of this function is shown in Figure 2.8.2.

Figure 2.8.2

Implicit differentiation can even be applied to an equation that does not by itself determine y as a function x. Sometimes extra inequalities must be assumed in order to make y a function of x.

EXAMPLE  4  Given

(2)

      find dy/dx. This equation does not determine y as a function of x; its graph is the unit circle. Nevertheless we differentiate both sides with respect to x and solve for dy/dx.

We can conclude that for any system of formulas S which contains the Equation 2 and also determines y as a function of x, it is true that

(3)   

We can use Equation 3 to find the slope of the line tangent to the unit circle at any point on the circle. The following examples are illustrated in Figure 2.8.3.

Figure 2.8.3

The system of formulas

x² + y² = 1,     y≥ 0

gives us                  

On the other hand the system

x² + y² = 1,     y ≤ 0

gives us

EXAMPLE  5  Find the slope of the line tangent to the curve

(4)  x5y3 + xy6 = y +1

at the points (1,1), (1,-1), and (0, -1).

The three points are all on the curve, and the first two points have the same x coordinate, so Equation 4 does not by itself determine y as a function of x. We differentiate with respect to x,

and then solve for dy/dx,

(5)

Substituting,   

Equation 5 for dy/dx is true of any system S of formulas which contains

Equation 4 and determines y as a function of x.

Here is what generally happens in the method of implicit differentiation. Given an equation

(6)  ι (x, y) = σ (x, y)

between two terms which may involve the variables x and y, we differentiate both sides of the equation and obtain

(7)               

We then solve Equation 7 to get dy/dx equal to a term which typically involves both x and y. We can conclude that for any system of formulas which contains Equation 6 and determines y as a function of x, Equation 7 is true. Also, Equation 7 can be used to find the slope of the tangent line at any point on the curve ι (x, y) = σ (x, y).

PROBLEMS  FOR  SECTION 2.8

In Problems 1-26, find dy/dx by implicit differentiation. The answer may involve both x and y.

1  xy = 1                              2   2x² - 3y² = 4 ,   y ≤ 0

3  x³ + y³  = 2                         4   x ³ = y 5

5  y= 1/(x+y)                          6   y²+3y - 5 = x

7  x -² + y-² = 1                        8  xy ³  = y + x

9  x²+3xy+y² = 0                      10  x/y + 3y =2

11  x5 = y2 - y + 1                       12  ___________

13 ___________                       14  x4 + y4 =5

15 xy2 - 3x2 y + x = 1                   16  2xy-2 + x -2 = y

17 y= sin (xy)                         18  y=cos (x+y)

19 x= cos2 y                          20  x= sin y +cos y

21 y=e x+2y                                      22  e y =x2+ y

23 e x =1n y                           24   1n y = sin x 

25 y 2 = 1n (2x + 3y)                    26   1n (cos y) = 2x + 5 

In Problems 27-33, find the slope of the line tangent to the given curve at the given point or points.

27   x ² + xy +y ² = 7 at (1, 2) , and (-1, 3)

28   x - y3 = y at (0, 0) , (0, 1), (0, -1), (-6,2)

29   x² - y² = 3 at (2, 1) , (2,-1), (____ ,0)

30   tan y= x ² at (π/4, 1)

31   2sin² x = 3 cos y at ( π /3, π /3)

32   y+ e y = 1+ 1n x at (1,0)

33   e sin x = 1n y at (0, e)

34    Given the equation x² + y² = 1, find dy/dx and d²y/dx².

35    Given the equation 2x² - y² = 1, find dy/dx and d²y/dx².

36    Differentiating the equation x² = y² implicitly, we get dy/dx = x/y. This is undefined at the

      point (0, 0). Sketch the graph of the equation to see what happens at the point (0,0).

EXTRA  PROBLEMS FOR CHAPTER 2

1 Find the derivative of  f(x) = 4x3 - 2x + 1

2 Find the derivative of  f(t) = __________.

3 Find the slope of the curve y = x( 2x + 4) at the point (1, 6).

4  A particle moves according to the equation y= 1/(t ² - 4). Find the velocity as a function of t.

5  Given y= 1/x³ , express Δy and dy as functions of x and Δx.

6  Given y= 1/____ , express Δy and dy as functions of x and Δx.

7   Find d (x² + 1/x² ).

8   Find d (x - 1/x).

9   Find the equation of the line tangent to the curve y= 1/(x-2) at the point (1, -1).

10  Find the equation of the line tangent to the curve y= 1+______ at the point (1, 2).

11  Find dy/dx where y= -3x³ -5x+2.

12  Find dy/dx where y= (2x -5)-2.

13  Find ds/dt where s= (3t +4)(t ² -5).

14  Find ds/dt where s= (4t² -6)-1 + (1 -2t) -2.

15  Find du/dv where u= (2v² -5v + 1)/ (v3 - 4).

16  Find du/dv where u= (v +(1/v)) / (v- (1/v)).

17  Find dy/dx where y= x1/ 2 + 4x 3/2.

18  find dy/ dx where y= ________.

19  find dy/ dx where y= x 1/3 + x -1/4.

20  find dy/ dx where y=ex cos2 x.

21  find dy/ dx where x= ________, y > 0.

22  find dy/ dx where x= y -1/2 + y-1, y > 0.

23  find dy/ dx where y= _____________.

24  find dy/ dx where y= sina (2+______).

25  find dy/ dx where y= u-1/2, u=5x+4.

26  find dy/ dx where y= u5, u=2 - x3.

27  find the slope dy/ dx of the path of a particle moving so that y= 3t + ____, x = (1/t) - t².

28  find the slope dy/ dx of the path of a particle moving so that y= ____, x = __________.

29  find d² y/ dx² where y= __________

30  find d² y/ dx² where y= x/(x² + 2).

31  an object moves so that s= _______. Find the velocity v=ds/dt and the acceleration a= d²s/dt².

32  find dy/dx by implicit differentiation when x + y + 2x² +3y³ = 2.

33  find dy/dx by implicit differentiation when 3xy³ + 2x³y = 1.

34  find the slope of the line tangent to the curve ____________.

□35 find the derivative of f(x) = |x²-1|

□36 find the derivative of function

□37 let f(x) = (x-c)4/3 . Show that f ′(x) exists for all real x but that f ′′(c) does not exist.

□38 let n be a positive integer and c a real number. Show that there is a function g(x) which has an

nth derivative at x= c but does not have an (n+1) st derivative at x= c. That is, g(n) (c) exists

     but g(n+1) (c) does not.

□39 (a) let u= x, y=u². Show that at x = 0, dy/dx exists even though du/dx does not.

(b) let u=x4, y= |u|. Show that at x= 0, dy/dx exists even though dy/du does not.

□40 suppose g(x) is differentiable at x= c and f(x) = |g(x)|. Show that

(a) f ′(c) = g′(c) if g(c) >0,

(b) f ′(c) =- g′(c) if g(c) >0,

(c) f ′(c) =0 if g(c)=0 and g′(c)=0

(d) f ′(c) does not exist if g(c)=0 and g′(c)≠0

□41 Prove by induction that for every positive integer n, n<2n.

□42 Prove by induction that the sum of the first n odd positive integers is equal to n2,

       1+3+5+ …+（2n -1) = n2