Mathematical methods for economic theory

Martin J. Osborne
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2.3 Exercises on derivatives of functions defined implicitly

  1. Suppose that 2x2 + 6xy + y2 = c for some constant c. Find dy/dx.

    Solution

    dy/dx = −(2x + 3y)/(3x + y).

  2. Suppose that the functions f and g are differentiable and g(f(x)) = x for all values of x. Use implicit differentiation to find an expression for the derivative f'(x) in terms of the derivative of g.

    Solution

    g'(f(x))f'(x) = 1, so f'(x) = 1/g'(f(x)).

  3. The demand and the supply for a good both depend upon the price p of the good and the tax rate t: D = f(p,t) and S = g(p,t). For any given value of t, an equilibrium price is a solution of the equation f(p,t) = g(p,t). Assume that this equation defines p as a differentiable function of t. Find ∂p/∂t in terms of the partial derivatives of f and g.

    Solution

    We have f'1(p,t)(∂p/∂t) + f'2(p,t) = g'1(p,t)(∂p/∂t) + g'2(p,t), so
    ∂p
    ∂t
    		 =
    g'2(p,t)−f'2(p,t)
    f'1(p,t) − g'1(p,t)
    		.

  4. The demand for a good both depends upon the price p of the good and the tax rate t: D = f(p,h(t)). The supply of the good depends on the price: S = g(p). For any given value of t, an equilibrium price is a solution of the equation f(p,h(t)) = g(p). Assume that this equation defines p as a differentiable function of t. Find ∂p/∂t in terms of the derivatives of f, g, and h.

    Solution

    We have
    ∂p
    ∂t
    		 = 
    f2'(p,h(t))h'(t)
    g'(p)−f1'(p,h(t))
    		.

  5. Let f(x, y) = 2x2 + xy + y2.
    1. Find the equation of the tangent at (x, y) = (2, 0) to the level curve of f that passes through this point.
    2. Find the points at which the slope of the level curve for the value 8 is 0.

    Solution

    1. The slope at (x, y) of the level curve through (x, y) is −f'x(x, y)/f'y(x, y) = −(4x + y)/(x + 2y). Thus the slope of the tangent at (2, 0) to the level curve through (2, 0) is −8/2 = −4. The equation of the straight line through an arbitrary point (x0, y0) with slope m is y − y0 = m(x − x0), so the equation of the tangent at (2, 0) to the level curve that passes through this point is y = −4(x − 2) or y = −4x + 8.
    2. The slope of a level curve is 0 if and only if 4x + y = 0. The point (x, y) is on the level curve for the value 8 if 2x2 + xy + y2 = 8. The two equations imply that 2x2 − 4x2 + 16x2 = 8, or 7x2 = 4, or x = ±2√7/7. Thus the points at which the slope of the level curve for the value 8 is 0 are (a, −4a) and (−a, 4a), where a = 2√7/7.

  6. Let D = f(r, P) be the demand for an agricultural product when the price is P and the producers' total advertising expenditure is r; f is decreasing in P. Let S = g(w, P) be the supply, where w is an index of how favorable the weather has been; g is increasing in P. Assume that g'w(w, P) > 0. An equilibrium price satisfies f(r, P) = g(w, P); assume that this equation defines P implicitly as a differentiable function of r and w. Find ∂P/∂w and determine its sign.

    Solution

    We have
    ∂P
    ∂w
    		 = −
    g'w(w, P)
    g'P(w, P) − f'P(r, P)
    		 < 0.
    (The sign follows from f'P < 0, g'P > 0, and g'w > 0.)

  7. The equilibrium value of the variable x is the solution of the equation
    f(x, α, β) + g(h(x), k(α)) = 0,
    where α and β are parameters and f, g, h, and k are differentiable functions. How is the equilibrium value of x affected by a change in the parameter α (holding β constant)?

    Solution

    Differentiating with respect to α we obtain
    f'1(x, α, β)
    ∂x
    ∂α
    		 + f'2(x, α, β) + g'1(h(x), k(α))h'(x)
    ∂x
    ∂α
    		 + g2'(h(x), k(α))k'(α) = 0,
    so that
    ∂x
    ∂α
    		 = 
    −f'2(x, α, β) − g'2(h(x), k(α))k'(α)
    f'1(x, α, β) + g'1(h(x), k(α))h'(x)

  8. The equilibrium value of the variable x is the solution of the equation
    f(x, g(x, α), β) + h(x, β) = 0,
    where α and β are parameters and f, g, and h are differentiable functions. How is the equilibrium value of x affected by a change in the parameter α (holding β constant)?

    Solution

    Differentiating with respect to α we obtain
    f'1(x, g(x, α), β)
    ∂x
    ∂α
    		 + f'2(x, g(x, α), β) g'1(x, α)
    ∂x
    ∂α
    		 +
                      f'2(x, g(x, α), β) g'2(x, α) + h1'(x, β)
    ∂x
    ∂α
    		 = 0,
    so that
    ∂x
    ∂α
    		 =
    −f'2(x, g(x, α), β) g'2(x, α)
    f'1(x, g(x, α), β) + f'2(x, g(x, α), β) g'1(x, α) + h'1(x, β)
    		.

  9. The equilibrium value of the variable x depends on the parameters (a1, ..., an):
    f(a1, ..., an, x) = 0.
    Find the rate of change of x with respect to ai for any i = 1, ..., n.

    Solution

    The equation defines x implicitly as a function of ai (and all the other parameters). Differentiating it with respect to ai we get
    f'i(a1, ..., an, x) + f'n+1(a1, ..., an, x)(∂x/∂ai) = 0,
    so that
    ∂x/∂ai = −f'i(a1, ..., an, x)/f'n+1(a1, ..., an, x),
    or, in a different notation,
    ∂x
    ∂ai
    		 = −
    ∂f/∂ai
    ∂f/∂x
    		.

  10. The value of y is determined as a function of t by the equation
    ∫1
    t    f(x,y)dx = 0,
    where f is a differentiable function. Find dy/dt.

    Solution

    Differentiating with respect to t we have
    −f(t, y) + ∫1
    t    f'2(x, y)y'(t)dx = 0,
    so that
    y'(t) =
    f(t, y)
    ∫1
    t    f'2(x, y)dx
    		.

  11. The function g is defined implicitly by the condition F(f(x,y),g(y)) = h(y). Find the derivative g'(y) in terms of the functions F, f, g, and h and their derivatives.

    Solution

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

  12. Suppose that x is implicitly defined as a function of the parameter t by the equation f(x)/f'(x) − x = t, where f is a twice-differentiable function. Find dx/dt.

    Solution

    Differentiate the equation defining x with respect to t:
    [(f'(x))2 − f(x)f"(x)]/[(f'(x))2]dx/dt − dx/dt = 1,
    so that
    dx/dt = −[f'(x)]2/[f(x)f"(x)].