Problem Set 4 Answers
1. Derive the first-order partial derivatives of the following functions.
(a) ¦(x) º 12x
¦¢(x) º 12
(b) ¦(x) º 2x2
¦¢(x) º 4x
(c) ¦(x) º x + x2 + x3
¦¢(x) º 1 + 2x + 3x2
(d) ¦(x) º y + 3x2 + v3
¦¢(x) º 6x
(e) ¦(y) º axyb + 3z2
¦¢(x) º abxyb-1
2. By applying the product and quotient rules, derive the first-order partial derivatives of the following functions.
(a) ¦(x) º º
¦¢(x) º = = 0
(b) ¦(x) º º
¦¢(x) º = = 0
(c) ¦(x) º º
¦¢(x) º = =
=
=
(d) ¦(x) º (100-x)x º h(x) g(x)
¦¢(x) º h¢(x)g(x) + h(x)g¢(x) = (-1)x + (100-x)(1) = 100 – 2x
(e) ¦(y) º p(y) y º h(y) g(y)
¦¢(y) º h¢(y)g(y) + h(y)g¢(y) = p¢(y) y + p(y)(1) = p¢(y) y + p(y)
3. Apply the Nth-order derivative rule for locating the relative extrema of the following functions.
(a) ¦(x) º x2
¦¢(x) º ¦(1)(x) º 2x
Solving the equation ¦¢(x) = 0 we find that x = 0 is the solution.
Hence, xo = 0 is the stationary point.
Continuing, ¦¢¢(xo) º ¦(2)(xo) = 2.
Because 2 ¹ 0, the first non-zero derivative at the stationary point is the second derivative.
Because the second derivative is an even-numbered derivative, the stationary point is either a minimum or a maximum.
Because the first non-zero derivative evaluated at the stationary point is positive, that is, because ¦¢¢(xo) º ¦(2)(xo) = 2 > 0, the stationary point is a relative minimum.
(b) ¦(x) º -x2
¦¢(x) º ¦(1)(x) º -2x
Solving the equation ¦¢(x) = 0 we find that x = 0 is the solution.
Hence, xo = 0 is the stationary point.
Continuing, ¦¢¢(xo) º ¦(2)(xo) = -2.
Because the second derivative is an even-numbered derivative, the stationary point is either a minimum or a maximum.
Because the first non-zero derivative evaluated at the stationary point is negative, that is, because ¦¢¢(xo) º ¦(2)(xo) = -2 < 0, the stationary point is a relative maximum.
(c) ¦(x) º x3
¦¢(x) º ¦(1)(x) º 3x2
Solving the equation ¦¢(x) = 0 we find that x = 0 is the solution.
Hence, xo = 0 is the stationary point.
Continuing, ¦¢¢(xo) º ¦(2)(xo) º 6xo = 0.
Continuing, ¦¢¢¢(xo) º ¦(3)(xo) º 6 = 6 ¹ 0.
Because the third derivative is an odd-numbered derivative, the stationary point is a point of inflection.
(d) ¦(x) º x2 – 1/3 x3
¦¢(x) º ¦(1)(x) º 2x – x2 = x(2-x)
Solving the equation ¦¢(x) = 0 we find that, in this case, there are actually two solutions. One solution is x = 0 and the other solution is x = 2. Now we must evaluate the function at each of the two points. For this purpose, denote the first stationary point by xo = 0 and denote the second stationary point by xa = 2. At the first solution, we have the following derivative.
¦¢¢(xo) º ¦(2)(xo) º 2-2xo = 2-2(0) = 2 > 0.
Hence, the first non-zero derivative at the point xo = 0 is the second derivative, which is an even-numbered derivative. Hence, at xo = 0, the function is either a minimum or a maximum, but not a point of inflection. Because the first nonzero derivative at the stationary point is ¦(2)(xo) = 2 > 0, a positive value, the stationary point is a relative minimum.
At the second stationary point xa = 2, we need, once again, to evaluate derivatives.
Hence, ¦¢¢(xa) º ¦(2)(xa) º 2-2xa = 2-2(2) = -2 < 0, and the stationary point is, therefore, a relative maximum.
(e) ¦(x) º (5/3)x3 – x5
¦¢(x) º ¦(1)(x) º 5x2 – 5x4 = 5x2(1-x2).
The solutions to the equation ¦¢(x) = 0 are three, xa = 0, xb = +1 and xc = -1.
We need to evaluate the rule at each of the three points. But, first, take the successive derivatives.
¦¢¢(x) º ¦(2)(x) º 10x – 20x3 = 10x(1-2x2).
At xa = 0, the derivative is equal to zero; but at xb = +1 and xc = -1 the derivatives are nonzero. In particular, ¦(2)(xb) = 10(1)(1-2(1)2) = -10 < 0 and ¦(2)(xc) = 10(-1)(1-2(-1)2) = +10 > 0. Hence, at xb = +1 the stationary point is a relative maximum and at xc = -1 the stationary point is a relative minimum.
But, we still have to find the first non-zero derivative for the function at the point xa = 0. Continuing,
¦¢¢¢(xa) º ¦(3)(xa) º 10 – 60xa2 = 10 > 0.
Hence, at the stationary point xa = 0, the function possesses a point of inflection.
4. Find the least squares estimator of the parameter m in the regression model
yi = m + ei, i = 1, 2, .., N.
Let denote the estimator that you are trying to derive. It follows that the estimated regression model (once we have the estimator ) will be,
yi = + ei, i = 1, 2, .., N.
By definition, the sum of squared deviations about the regression line yi = is
SS º ,
or
SS() º .
This is just a sum of squared terms about the regression mean—that is, a quadratic function in the variable —and we can apply the (now familiar) calculus in order
to derive the estimator.
The first-order partial derivative is
SS¢() º = .
Locating the stationary point SS¢() = 0, we solve
= = 0.
Solving, we obtain
.
In other words, the estimator that minimizes the sum of squared deviations about the regression line is the sample mean.