Mathematical methods for economic theory

Martin J. Osborne
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1.5 Exercises on calculus: one variable

  1. Find the first derivatives of
    1. 2x4 + 3x1/2 + 7
    2. 4/x2
    3. (x + 3)/(x2 + x)
    4. xln x

    Solution

    1. 8x3 + (3/2)x−1/2
    2. −8/x3
    3. (−x2 − 6x − 3)/(x2 + x)2
    4. ln x + 1

  2. Show that for any values of a and b for which ab > 0, the derivative of the function ax/(ax + b(1 − x)) with respect to x is positive. For a = 1 and b = 2, calculate f(1) and f(3). How are these two results consistent? (Is the derivative really positive for all values of x?)

    Solution

    The derivative is ab/(ax + b(1 − x))2. For a = 1 and b = 2, the value of the function at x = 1 is 1 and its value at x = 3 is −3. The function is discontinuous at x = 2. At every point x ≠ 2 the derivative is positive, but at x = 2 it is not defined (the value of the function is very large and positive for values of x a bit smaller than 2 and is very large and negative for values of x a bit larger than 2).

  3. At what points is the function |x| differentiable?

    Solution

    The function is differentiable at all points except 0.

  4. Find the following indefinite integrals, remembering that the derivative of eax is aeax (an implication of the chain rule, studied in a later section).
    1. ∫x4dx
    2. ∫2e−2xdx
    3. ∫x(1 + x)1/2dx (use integration by parts)
    4. ∫(4x + 2)/(x2 + x)dx

    Solution

    1. (1/5)x5 + c
    2. −e−2x + c
    3. x(2/3)(1 + x)3/2 − (4/15)(1 + x)5/2 + c = (2/15)(−2 + 3x)(1 + x)3/2 + c. (You can alternatively use the fact that ∫x(1 + x)1/2dx = ∫((1 + x)(1 + x)1/2 − (1 + x)1/2)dx = ∫(1 + x)3/2dx − ∫(1 + x)1/2dx to calculate the integral without using integration by parts.)
    4. 2ln |x2 + x| + c

  5. Find the following definite integrals.
    1. ∫3
      1      3x1/2dx
    2. ∫3
      2      (e2x + ex)dx
    3. ∫∞
      0      e−rtdt where r > 0 is a constant

    Solution

    1. 2(33/2−1)
    2. (1/2)e6 + e3 − (1/2)e4 − e2
    3. 1/r. (Note that because r > 0, e−rt converges to 0 as t increases without bound.)