Mathematical methods for economic theory

Martin J. Osborne
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2.2 Exercises on the chain rule

  1. Find the derivatives of the following functions of a single variable.
    1. f(x) = (3x2 − 1)3.
    2. f(x) = xe2x. (Remember that ln(ax) = xln a, so that ax = exln a.)
    3. f(x) = 2x + x2.
    4. f(x) = ln x2.
    5. f(x) = sin bx, where b is a constant.

    Solution

    1. f'(x) = 18x(3x2 − 1)2.
    2. f'(x) = e2x + x2x(ln 2)e2x.
    3. f'(x) = 2xln 2 + 2x.
    4. f'(x) = 2/x.
    5. f'(x) = bcos bx.

  2. As in an example in the text, consider a profit-maximizing firm that produces a single output with a single input. Denote its (differentiable) production function by f, the price of the input by w, and the price of the output by p. Suppose that its profit-maximizing input when the prices are w and p is z(w, p), so that its maximized profit is
    Π(w, p) = pf(z(w, p)) − wz(w, p).
    Find the derivative of Π with respect to w.

    Solution

    Using the chain rule we have
    Π'w(w, p)  =  pf'(z(w, p))z'w(w, p) − wz'w(w, p) − z(w, p)
    or
    Π'w(w, p)  =  z'w(w, p)[pf'(z(w, p)) − w] − z(w, p).
    But if z(w, p) > 0 then pf'(z(w, p)) − w = 0, the “first-order condition” for maximal profit. Thus if z(w, p) > 0 then we have
    Π'w(w, p) = −z(w, p).

  3. Define the function F of two variables by F(x, y) = f(g(x, y), h(x, y)) for all (x, y), where f(s, t) = st2, g(x, y) = x + y2, and h(x, y) = x2y. Use the chain rule to find F'x(x, y) and F'y(x, y).

    Solution

    We have F'x(x, y) = f'1(g(x, y), h(x, y))g'x(x, y) + f'2(g(x, y), h(x, y))h'x(x, y), f'1(s, t) = t2, g'x(x, y) = 1, f'2(s, t) = 2st, and h'x(x, y) = 2xy. Thus F'x(x, y) = (x2y)2 + 2(x + y2)x2y(2xy) = x4y2 + 4x4y2 + 4x3y4 = 5x4y2 + 4x3y4. Similarly, F'y(x, y) = 2x5y + 4x4y3.

  4. Define the function F of two variables by F(x, y) = f(g(x, y), h(k(x))), where f, g, h, and k are differentiable functions. Find the partial derivative of F with respect to x in terms of the partial derivatives of f, g, h, and k.

    Solution

    We have F'x(x, y) = f'1(g(x, y), h(k(x)))g'1(x, y) + f'2(g(x, y), h(k(x)))h'(k(x))k'(x).

  5. Define the function F of two variables by F(p, q) = pf(p, q, m(p, q)), where f and m are differentiable functions. Find an expression for the partial derivative of F with respect to p in terms of the partial derivatives of f and m.

    Solution

    We have F'p(p, q) = f(p, q, m(p, q)) + p[f'1(p, q, m(p, q)) + f'3(p, q, m(p, q))m'1(p, q)].

  6. Define the function F of two variables by F(x, y) = h(f(x), g(x, y)), where f, g, and h are differentiable functions. Find the derivative of F with respect to x in terms of the partial derivatives of f, g, and h.

    Solution

    We have F'x(x, y) = h'1(f(x), g(x, y))f'(x) + h'2(f(x), g(x, y))g'1(x, y).

  7. Define the functions U and V of two variables by U(x, y) = F(f(x) + g(y)) for all (x, y), and V(x, y) = ln[U'x(x, y)/U'y(x, y)] for all (x, y), where f, g, and F are twice-differentiable functions. Find V"xy(x, y).

    Solution

    We have U'x(x, y) = F'(f(x) + g(y))f'(x) and U'y(x, y) = F'(f(x) + g(y))g'(y). Thus V(x, y) = ln f'(x) − ln g'(y).

    Hence V'x(x, y) = f"(x)/f'(x), and thus V"xy(x, y) = 0 for all (x, y).

  8. Let
    y = F(x1(p), ..., xn(p), p) − p·x(p),
    where F and xi for i = 1, ..., n are differentiable functions, p is an n-vector, x(p) denotes the vector (x1(p), ..., xn(p)), and p·x(p) denotes the inner product of p and x(p). Find the derivative of y with respect to pi, given pj for j ≠ i. (Use the notation ∂xj/∂pi for the partial derivative of xj with respect to pi.)

    Solution

    The function F has 2n arguments: x1(p), ..., xn(p), and p1, ..., pn. In applying the chain rule, we differentiate F with respect to each of these arguments, and then differentiate each argument with respect to pi. The derivative of each of the last n arguments except pi (which is the (n + i)th argument of F) with respect to pi is zero. Thus the derivative of F(x1(p), ..., xn(p), p) with respect to pi is
    ∑n
    j=1    F'j(x1(p), ..., xn(p), p)(∂xj/∂pi) + F'n+i(x1(p), ..., xn(p), p).
    Now consider the last term, −p·x(p). We may write this term as −∑n
    j=1    pjxj(p). When we differentiate the jth term in this summation, for j ≠ i, with respect to pi we obtain −pj(∂xj/∂pi); when we differentiate the ith term with respect to pi we obtain −xi(p) − pj(∂xj/∂pi). In summary, we have
    ∂y/∂pi  =  ∑n
    j=1    F'j(x1(p), ..., xn(p), p)(∂xj/∂pi)     + F'n+i(x1(p), ..., xn(p), p) − xi(p) − ∑n
    j=1    pj(∂xj/∂pi)    .

  9. The amount x of some good demanded depends on the price p of the good and the amount a the producer spends on advertising: x = f(p, a), with f'p(p, a) < 0 and f'a(p, a) > 0 for all (p, a). The price depends on the weather, measured by the parameter w, and the tax rate t: p = g(w, t), where g'w(w, t) > 0 and g't(w, t) < 0 for all (w, t). The amount of advertising depends only on t: a = h(t), with h'(t) > 0. If the tax rate increases does the demand for the good necessarily increase or necessarily decrease, or neither?

    Solution

    We have x = f(g(w, t), h(t)), so ∂x/∂t = f'p(g(w, t), h(t)) · g't(w, t) + f'a(g(w, t), h(t)) · h'(t). Thus, given the signs of the derivatives specified in the question, demand increases as t increases.

  10. Let
    H(r) = ∫g(r)
    0     e−rtf(t) dt,
    where g and f are differentiable functions. Find H'(r).

    Solution

    H'(r) = e−rg(r)f(g(r))g'(r) − ∫g(r)
    0    te−rtf(t)dt.

  11. Let
    H(t) = ∫t
    t−T    g(x)e−δ(t−x) dx,
    where g is a differentiable function and δ and T are constants. Find the derivative H'(t).

    Solution

    We have
    H'(t)  = g(t)e−δ(t−t) − g(t − T)e−δ(t−t+T) − δ∫t
    t−T    g(x)e−δ(t−x) dx
    = g(t) − g(t − T)e−δT − δH(t).

  12. Suppose that the amount x of some good demanded depends on the price p of the good and the price q of another good; both these prices depend on two parameters, α and β (e.g. the weather, the rate of a government subsidy). You observe that ∂x/∂α > 0, ∂p/∂α < 0, ∂q/∂α > 0, and |∂p/∂α| > |∂q/∂α|. You have a theory that x = f(p(α,β),q(α,β)), where f'1(p,q) > 0 and f'2(p,q) > 0 for all (p,q). Are your observations consistent with your theory? Are your observations consistent with a theory that imposes the stronger restriction that f'1(p,q) > f'2(p,q) > 0 for all (p,q)?

    Solution

    We have
    ∂x
    ∂α
    		 = f'1(p(α, β), q(α, β))
    ∂p
    ∂α
    		 + f'2(p(α, β), q(α, β))
    ∂q
    ∂α
    		,
    so the observations that ∂x/∂α > 0, ∂p/∂α < 0, ∂q/∂α > 0, and |∂p/∂α| > |∂q/∂α| are consistent with f'1(p, q) > 0 and f'2(p, q) > 0 for all (p, q) (f'1(p, q) simply has to be small enough relative to f'2(p, q)). However, the observations are not consistent with f'1(p, q) > f'2(p, q) > 0 for all (p, q), since in this case ∂x/∂α < f'2(p, q)(∂p/∂α + ∂q/∂α) < 0 since |∂p/∂α| > |∂q/∂α| and f'2(p, q) > 0.

  13. A firm faces uncertain demand D and has existing inventory I. The firm wants to choose its stock level Q to minimize the value of the function
    g(Q) = c(Q − I) + h∫Q
    0    (Q−D)f(D)dD + p∫a
    Q    (D−Q)f(D)dD,
    where c, I, h, p, and a are positive constants with p > c, and f is a nonnegative function that satisfies ∫a
    0    f(D)dD     = 1 (so that it can be interpreted as a probability distribution function). (The first term represents the cost of the new stock; the second term represents the cost of overstocking; and the third term represents the cost of understocking (you miss sales, and the customers who are turned away may not come back in the future).)
    1. Find g'(Q) and g''(Q) and show that g''(Q) ≥ 0 for all Q.
    2. Define F(Q*) = ∫Q*
      0      f(D)dD, where Q* is the stock level that minimizes g(Q). Use the "first-order" condition g'(Q*) = 0 to find F(Q*) (the probability that demand D does not exceed Q*) in terms of the parameters p, c, and h. (Hint: Use the fact that ∫Q*
      0      f(D)dD + ∫a
      Q*      f(D)dD = ∫a
      0      f(D)dD = 1.)

    Solution

    1. Using Leibniz' rule, we have
      g'(Q) = c + h∫Q
      0      f(D)dD − p∫a
      Q      f(D)dD.
      To find the derivative of this expression, we use Leibniz' rule on each integral separately. The derivative of ∫Q
      0      f(D)dD is f(Q) (note that Q does not appear in the integrand, so that the last term in Leibniz' formula is 0) and the derivative of ∫a
      Q      f(D)dD is −f(Q). Thus
      g''(Q) = (h + p)f(Q)
      and hence g''(Q) ≥ 0 for all Q.
    2. The first-order condition g'(Q*) = 0 implies that
      c + h∫Q*
      0      f(D)dD − p∫a
      Q*      f(D)dD = 0.
      Now, F(Q*) = ∫Q*
      0      f(D)dD and
       ∫a
      Q*      f(D)dD = ∫a
      0      f(D)dD − ∫Q*
      0      f(D)dD = 1 − F(Q*),
      so that
      c + hF(Q*) − p(1 − F(Q*)) = 0,
      and hence F(Q*) = (p−c)/(h+p).