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David Terr
Ph.D. Math, UC Berkeley

 

Home >> Pre-Calculus >> 11. Introduction to Calculus

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11.4. The Quotient Rule

The last differentiation rule we will derive is the quotient rule, which will allow us to compute derivatives of rational functions. Before stating and proving this rule, however, we will find another rule useful, namely the rule for computing the derivative of the recipricol of a function.

 

Theorem 11.4.1: For differentiable function f(x), we have

recipricol rule

Proof: We apply the product rule to the product 1 = f(x) / f(x), obtaining the following:

proof of recipricol rule

By d ividing the right side by f(x) and subtracting -1 / f(x)2 df / dx from both sides, we obtain the statement of the theorem.

QED

 

Theorem 11.4.2 (Quotient Rule): Let f(x) and g(x) be differentiable functions. Then we have

quotient rule

Proof: We prove the quotient rule by applying the product rule to the product f(x) * 1/g(x), using the recipricol rule to compute d(1/g(x)) / dx. We have

. proof of quotient rule

QED

 

As mentioned earlier, the quotient rule may be used to compute the derivatives of rational functions, as the following example shows.

 

Example 1: Compute the derivatives of the following rational functions and evaluate them at x = 5.

  • (a) f(x) = 1 / (1 - x)
  • (b) f(x) = x / (1 - x)
  • (c) f(x) = x / (1 - x2)
  • (d) f(x) = (1 - x2) / (1 + x2)
  • (e) f(x) = (x2 - 4) / (3x2 + 8x - 3)

Solution:

  • (a) We have f(x) = p(x) / q(x), with p(x) = 1 and q(x) = 1 - x. By the quotient rule, we have

f'(x) = [q(x)p'(x) - p(x)q'(x)] / q(x)2

= [(1 - x)(0) - (1)(-1)] / (1 - x)2

= 1 / (1 - x)2.

Thus we have f'(5) = 1 / (1 - 5)2 = -1/16.

  • (b) We have f(x) = p(x) / q(x), with p(x) = x and q(x) = 1 - x. By the quotient rule, we have

f'(x) = [q(x)p'(x) - p(x)q'(x)] / q(x)2

= [(1 - x)(1) - (x)(-1)] / (1 - x)2

= (1 - x + x) / (1 - x)2

= 1 / (1 - x)2.

Thus we have f'(5) = 1 / (1 - 5)2 = -1/16.

  • (c) We have f(x) = p(x) / q(x), with p(x) = x and q(x) = 1 - x2. By the quotient rule, we have

f'(x) = [q(x)p'(x) - p(x)q'(x)] / q(x)2

= [(1 - x2)(1) - (x)(-2x)] / (1 - x2)2

= (1 - x2 + 2x2) / (1 - x2)2

= (1 + x2) / (1 - x2)2.

Thus we have f'(5) = (1 + 52) / (1 - 52)2 = 26 / (-24)2 = -26 / 576 = -13 / 288.

  • (d) We have f(x) = p(x) / q(x), with p(x) = 1 - x2 and q(x) = 1 + x2. By the quotient rule, we have

f'(x) = [q(x)p'(x) - p(x)q'(x)] / q(x)2

= [(1 + x2)(-2x) - (1 - x2)(2x)] / (1 + x2)2

= (-2x - 2x3 - 2x + 2x3) / (1 + x2)2

= -4x / (1 + x2)2.

Thus we have f'(5) = -(4)(5) / (1 + 52)2 = -20 / 262 = -20 / 576 = -5 / 144.

  • (e) We have f(x) = p(x) / q(x), with p(x) = x2 - 4 and q(x) = 3x2 + 8x - 3. By the quotient rule, we have

f'(x) = [q(x)p'(x) - p(x)q'(x)] / q(x)2

= [(3x2 + 8x - 3)(2x) - (x2 - 4)(6x - 8)] / (3x2 + 8x - 3)2

= (6x3 + 16x2 - 6x - 6x3 + 8x2 + 24x - 32) / (3x2 + 8x - 3)2

= (24x2 + 18x - 32) / (3x2 + 8x - 3)2.

Thus we have f'(5) =[(24)(52) + (18)(5) - 32]/ [(3)(52) + (8)(5) - 3]2 = 358 / 1122 = 179 / 6272.

 

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