2.3 DERIVATIVES OF RATIONAL FUNCTIONS

A term of the form

                    a1 x + a0

Where a1, a0are real numbers, is called a linear term in x; if a1≠ 0, it is also called polynomial of degree one in x. A term of the form

                    a2x²+ a1x + a0,        a2 ≠ 0

is called a polynomial of degreetwo in x, and, in general, a term of the form

                         anxn+ an-1xn-1+ …a1x+ a0,    an ≠0

is called a polynomial of degreeninx.

 

A rational term in x is any term which is built up from the variable x and real numbers using the operations of addition, multiplication, subtraction, and division. For example every polynomial is a rational term and so are the terms

 

                          

 

 

 

A linear function, polynomial function, or rational function is a function which is given by a linear term, polynomial, or rational term, respectively. In this section we shall establish a set of rules which enable us to quickly differentiate any rational function. The rules will also be useful later on in differentiating other functions.

 

THEOREM 1

      The derivative of a linear function is equal to the coefficient of x. That is,

                  

 

 

 

 

 

 

 

 

 

PROOF  Let y=bx + c, and let Δx≠ 0 be infinitesimal. Then

                 

               

                  

                       

 

 

Therefore   

 

Multiplying through by dx, we obtain at once

                        dy = bdx.

 

If in Theorem 1 we put b=1, c=0, we see that the derivative of the identity function f(x)=x is f '(x) = 1; i.e.,

                         

 

 

On the other hand, if we put b=0 in Theorem 1 then the term bx+ cis just the constant c, and we find that the derivative of the constant function f(x) =c is f '(x)= 0; i.e.,

                           

 

 

 

THEOREM 2  (Sum Rule)

 Suppose u and v depend on the independent variable x. Then for any value of x where du/ dx and dv/dx exist,

 

 

 

      In other words, the derivative of the sum is the sum of the derivatives.

 

PROOF  Let y=u+v, and let Δx≠0 be infinitesimal. Then

 

 

 

Taking standard parts,

                   

 

 

Thus                   

 

 

By using the Sum Rule n-1  times, we see that

 

 

 

THEOREM 3  (Constant  Rule)

       Suppose u depends on x, and c is a real number. Then for any value of x where du/dx exists,

                 

 

 

 

PROOF  Let y=cu, and let Δx≠0  be infinitesimal. Then

        

 

 

 

            

 

Taking standard parts,

                    

 

 

Whence              

 

 

 

The Constant Rule shows that in computing derivatives, a constant factor may be moved “outside” the derivative. It can only be used whenc is a constant. For products of two functions ofx, we have:

 

THEOREM  4 ( Product Rule)

   Suppose u and v depend on x. Then for any value of x where du/dx and dv/dx exist,

   

 

PROOF  Let y=uv, and let Δx≠0 be infinitesimal.

                

 

 

 

 

 

Δu is infinitesimal by the Increment Theorem, whence

                     

 

 

 

 

 

 

So       

 

 

 

The Constant Rule is really the special case of the Product Rule where v is a constant function of x , v= c. To check this we let v be the constant c and see what the Product Rule gives us:

 

 

 

 

This is the Constant Rule.

 

The Product Rule can also be used to find the derivative of a power of u.

 

THEOREM 5  (Power Rule)

      Let u depend on x and let n be a positive integer. For any value of x where du/dx exists,

 

 

 

 

PROOF  To see what is going on we first prove the Power Rule for n = 1,2,3,4.

         n =1:  We have u n = u and u0 = 1, whence 

              

 

 

          n =2:  We use the Product Rule,

 

 

          n =3:  We write u 3 = u ·u², use the Product Rule again, and then use the result   

                for n = 2.

 

 

        

 

          n =4:  Using the Product Rule and then the result for n = 3,

 

 

 

 

 

We can continue this process indefinitely and prove the theorem for every positive integer n. To see this , assume that we have proved the theorem for m. That is, assume that

 

(1)                    

 

 

   We then show that it is also true for m+1. Using the Product Rule and the Equation 1,

  

 

 

 

 

Thus         

 

 

 

This shows that the theorem holds for m+1.

We have shown the theorem is true for 1,2,3,4. Set m=4; then the theorem holds for m+1 = 5. Set m= 5; then it holds for m+1 = 6. And so on.

Hence the theorem is true for all positive integers n.

 

In the proof of the Power Ruler, we used the following principle:

 

PRINCIPLE  OF  INDUCTION

 

     Suppose a statement P (n) about an arbitrary integer n is true when n=1.

     Suppose further that for any positive integer m such that P(m) is true, P(m+ 1) is also true. Then the statement P(n) is true of every positive integer n.

 

In the previous proof, P (n) was the Power Rule,

              

 

 

The principle of induction can be made plausible in the following way. Let a positive integer n be given. Set m=1; since P (1) is true, P(2) is true. Now set m=2; since P(2) is true, P(3) is true. We continue reasoning in this way for n steps and conclude that P(n) is true.

 

The Power Rule also holds for n=0 because when u≠ 0  , u0  =1  and d1/dx = 0.

Using the Sum, Constant, and Power rules, we can compute the derivative of a polynomial function very easily. We have

 

 

 

 

 

and thus

EXAMPLE  1   

 

EXAMPLE  2    

 

 

Two useful facts can be stated as corollaries.

 

COROLLARY  1

 

          The derivative of a polynomial of degree n> 0 is a polynomial of degree n-1.

          (A nonzero constant is counted as a polynomial of degree zero.)

 

COROLLARY  2

If u depends on x, then    

 

Whenever du/dx exists. That is, adding a constant to a function does not change its derivative.

 

In Figure 2.3.1  

         we see that the effect of adding a constant is to move the curve up or down               the y-axis without changing the slope.

 

         For the last two rules in this section we need the formula for the derivative of 1/v.

 

 

 

 

 

 

 

 

 

 

 

Figure 2.3.1

 

LEMMA

     Suppose v depends on x. Then for any value of x where v≠ 0 and dv/dx exists,

 

 

 

PROOF  Let y=1/v and let Δx ≠ 0 be infinitesimal.

 

 

 

 

 

 

 

 

 

Taking standard parts,

 

 

 

 

 

 

 

 

 

Therefore        

 

 

 

THEOREM  6   (Quotient Rule)

 

     Suppose u, v depend on x. Then for any value of x where du/dx, dv/dx exist and v≠ 0,

 

 

 

 

PROOF  We combine the Product Rule and the formula for d(1/v). Let y= u/v.

         We write y in the form

 

 

Then

 

 

 

 

 

 

THEOREM  7 (Power Rule for Negative Exponents)

     Suppose u depends on x and n is a negative integer. Then for any value of x where du/dx exists and u≠ 0, d(un)/ dx exists and

 

 

PROOF  Since n is negative, n= -m where m is positive. Let y=un = u-m. Then y= 1/um. By 

         the lemma and the Power Rule,

 

 

 

 

 

 

 

 

 

 

The Quotient Rule together with the Constant, Sum, Product, and Power Rules make it easy to differentiate any rational function.

 

EXAMPLE 3   Find dy when

                       

 

     Introduce the new variable u with the equation

                              u=x² - 3x + 1.

     Then y= 1/u, and du = (2x-3)dx , so

                     

 

 

 

EXAMPLE  4  Let ___________ and find dy.

Let         u=(x4-2) 3, v=5x-1.

Then      

 

 

Also,        du= 3 ·(x4-2) 2 4x 3 dx = 12 (x4 - 2) 2 x3 dx,

            dv=5dx.

 

Therefore

 

 

 

 

EXAMPLE  5  Let y= 1/x3 + 3/x2 + 4/x +5.

  

    Then         

 

EXAMPLE  6  Find dy where

 

         

 

 

 

 

This problem can be worked by means of a double substitution. Let

                         

 

 

 

Then            y= v².

We find dy, dv and du,

                      dy= 2v dv,

                      dv= -u -2 du,

                      du= (2x+1) dx.

Substituting, we get dy in terms of x and dx,

 

 

 

 

 

 

EXAMPLE  7   Assume that u and v depend on x. Given y= (uv)-2, find dy/dx in terms of

       du/dx and dv/ dx.

        Let s =uv, whence y=s-2. We have

                    dy = -2 s -3 ds,

                    ds=udv + vdu.

        Substituting,  dy= -2(uv)-3 (udv+vdu),

      

        and        

 

 

 

         The six rules for differentiation which we have proved in this section are so useful that they should be memorized. We list them all together.

 

 

 

 

 

 

 

 

 

          Table 2.3.1   Rules for Differentiation

 

 

 

 

 

 

 

 

 

 

 

 

 

An easy way to remember the way the signs are in the Quotient Rule 6 is to put u=1 and use the Power Rule 5 with n= -1,

                  

 

 

 

 PROBLEMS  FOR  SECTION 2.3

In Problems 1-42 below, find the derivative.

1   f(x) = 3x2 + 5x -4                 2  _______________

3   y=(x+8) 5                               4  z = (2+3x)4

5  f(t) = (4-t)3                       6   g(x) = 3(2-5x)26         

7  y = (x2 + 5)3                      8    u = (6+2x2)3

9  u = (6-2x2)3                             10   w= (1+4x3) -2

11  w= (1 - 4x3) -2                         12   y= 1 + x-1 + x -2 +x -3

13  f(x) = 5(x+1-1/x)                 14    u = (x2 +3x +1)4

15  v= 4 (2x2 - x + 3) -2                   16    y= -(2x + 3 +4x -1) -1

17   ________________               18   ________________

19   ________________               20  s = (2t+1)(3t-2)

21   _______________                22  y= (2x3 +4 )(x 2 - 3x+1)

23   v= (3t2 +1 )(2t - 4)3                         24  z= ( -2x +4 +3x -1)( x+1 - 5x-1)

25   ________________               26  ___________________

27   ________________               28  ____________________

29   ________________               30  ____________________

31   ________________               32  y = 4x-5

33   y=6                            34  y= 2x( 3x-1 ) (4-2x )

35   y= 3(x2 +1 )(2x 2 - 1)(2x+3)         36  y= (4x +3 )-1 + (x - 4)-2

 

 

37   _________________      38   y = (x2 +1 )-1 (3x -1)-2

39   y= [(2x +1 )-1 +3]-1          40   s = [(t 2 +1 )3 +t]-1 

41   y= (2x +1) 3 (x2+1]2         42   _________________

 

In Problems 43-48, assume u and v depend on x and find dy/dx in terms of du/dx and dv/dx.

43  y = u-v                 44     y=u2v

45  y = 4u + v2                  46     y = 1/(u+v)

47   y = 1/uv               48     y=(u+v) (2u - v)

49   Find the line tangent to the curve y=1+x+x² + x³ at the point (1,4).

50   Find the line tangent to the curve y=9x-² at the point (3,1).

□51  Consider the parabola y=x² + bx +c. Find values of b and c such that the line y=2x is

      tangent to the parabola at the point x=2, y=4.

□52  Show that if u, v, and w are differentiable functions of x and y=ucw, then

 

 

 

□53   Use the principle of induction to show that if n is a positive integer, u1,…… un are

       differentiable functions of x, and y=u1 + un, then

 

 

□54   Use the principle of induction to prove that for every positive integer n,

 

 

□55 Every rational function can be written as a quotient of two polynomials, p(x)/q(x).

     Using this fact, show that the derivative of every rational function is a rational

      Function.