Exercises 3

This page allows you to practice some exercises on Differentiation. If you notice that you have difficulties, we advise you to go over the corresponding sections in the book.

These questions correspond to the following sections: - Section 6.8 - The Chain Rule - Section 6.9 - Higher-Order Derivatives

Question 16

Use the chain rule to find $dz/dx$ for the following:

(a) $z = 5y^4,$ where $y = 1 + x^2$

(b) $z = y − y^6,$ where $y = 1 + \frac{1}{x}$

Show answer

(a) $\frac{\text{d}z}{\text{d}x} = \frac{\text{d}z}{\text{d}y} \cdot \frac{\text{d}y}{\text{d}x} = 20 y^{3} \text{d}y/\text{d}x = 20 (1+x^{2})^{3} 2x = 40x(1+x^{2})^{3}$.

(b) $\frac{\text{d}z}{\text{d}x} = \frac{\text{d}z}{\text{d}y} \cdot \frac{\text{d}y}{\text{d}x} = (1 - 6y^{5}) \text{d}y/\text{d}x =(1 - 6(1 + 1/x)^{5}) \cdot (-1/x^{2}) = - \frac{1}{x^{2}(1-6(1+1/x)^{5})}$.

Question 17

Find the derivatives of the following functions:

(a) $y = \frac{1}{(x^2 + x + 1)^5}$

(b) $y = \sqrt{x + \sqrt{x + \sqrt{x}}}$

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(a) $y' = -5(x^{2} + x + 1)^{-6} (2x + 1)$.

(b) $y' = \frac{1}{2}\left[ x + (x + x^{1/2})^{1/2} \right]^{-1/2} \left( 1 + \frac{1}{2}(x + x^{1/2})^{-1/2} (1 + \frac{1}{2}x^{-1/2} ) \right)$.

Given the odd form of the function and its derivative you might wonder what it looks like in practice. Below you can find a graph of $y$ and its derivative $y'$. We can see that $y$ increases smoothly as $x$ increases, which is also reflected in $y'$. The derivative is strictly positive but rapidly becomes smaller in magnitude, which indicates that $y$ is increasing sharply at first, but continues to grow more slowly as $x$ gets larger.

Question 18

Consider the demand function $x = b − \sqrt{ap − c}$ , where $a, b,$ and $c$ are positive constants, $x$ is the quantity demanded, and the price $p$ satisfies $p > c/a$. Compute $dx/dp$.

Show answer

We define $x = b − \sqrt{ap − c} = b − \sqrt{u}$, with $u = ap − c$. Then $\frac{\text{d}x}{\text{d}p} = − \frac{1}{2\sqrt{u}}u' = − \frac{a}{2\sqrt{ap − c}}$.

Question 19

Find a formula for $h'(x)$ when:

(a) $h(x) = f (x^2)$

(b) $h(x) = g(x^n f (x))$

Show answer

(a) $h'(x) = f'(x^{2}) 2x$, by using the chain rule.

(b) $h'(x) = g'(x^{n}f(x)) \cdot \left[x^{n} f'(x) + nx^{n-1}f(x) \right]$, by using both the product and chain rule.

Question 20

Differentiate each of the following in two different ways:

(a) $y = (x^4)^5 = x^{20}$

(b) $y = (1 − x)^3 = 1 − 3x + 3x^2 − x^3$

Show answer

(a) You can directly take the derivative of $y = x^{20}$, which yields $y'= 20x^{19}$. Alternatively, you can apply the chain rule to the first expression. So using $y = (x^{4})^{5}$ , we obtain the result $y' = 5(x^{4})^{4} \cdot 4x^{3} = 20x^{19}$.

(b) We again start with the expression on the right. The derivative can easily be obtained as $y' = -3 + 6x - 3x^{2}$. Using the first expression, $y = (1- x)^{3}$, we have to use the chain rule and obtain $y' = 3(1-x)^{2} \cdot -1 = -3(1 - 2x + x^{2}) = -3 + 6x - 3x^{2}$.

Question 21

If $f$ is differentiable at $x$, find expressions for the derivatives of the following functions:

(a) $[f (x)]^2 − x$

(b) $x^2 f (x) + [f (x)]^3$

(c) $\sqrt{f (x)}$

(d) $x^2 / f(x)$

(e) $[f (x)]^2/x^3$

Show answer

(a) $2f(x)f'(x) - 1$.

(b) $2x f(x) + x^{2} f'(x) + 3[f(x)]^{2} f'(x)$.

(c) $\frac{1}{2} [f(x)]^{-1/2} \cdot f'(x) = f'(x)/[2\sqrt{f(x)}]$.

(d) $[2x f(x) - x^{2} f'(x)]/[f(x)]^{2}$ using the quotient rule.

(e) $[x^{3} 2 f(x) f'(x) - f(x)^{2} 3x^{2}]/[x^{3}]^{2} = [2x^{3} f(x) f'(x) - 3x^{2}f(x)^{2}]/x^{6}$.

Question 22

Compute the second derivatives of:

(a) $y = x^5 − 3x^4 + 2$

(b) $y = (1 + x^2)^{10}$

(c) $y = \sqrt{(1 + x^2)} = (1 + x^2)^{1/2}$

(d) $y = 3x^3 + 2x − 1$

Show answer

(a) Note that $y' = 5x^{4} - 12x^{3}$, so $y'' = 20x^{3} - 36x^{2}$.

(b) Note that $y' = 10(1+x^{2})^{9} \cdot 2x = 20x(1+x^{2})^{9}$. Now we obtain for the second derivative $y'' = 20(1 + x^{2})^{9} + 20x \cdot 9 \cdot (1+ x^{2})^{8} \cdot 2x = 20 (1+x^{2})^{8}(1 + 19x^{2})$.

(c) Note that $y' = \frac{1}{2} (1 + x^{2})^{-1/2} \cdot 2x = x(1+x^{2})^{-1/2}$. The second derivative is given by $y'' = (1 + x^{2})^{-1/2} + x \cdot -\frac{1}{2}(1+x^{2})^{-3/2} \cdot 2x = (1+x^{2})^{-3/2}$.

(d) $y' = 9x^{2} + 2$, so $y'' = 18x$.

Question 23

For the following functions, compute the higher-order derivative as indicated in the exercise:

(a) $d^3 z/dt^3$ for $z = 120t − (1/3)t^3$

(b) $f^{(4)} (1)$ for $f(z) = 100z^{−4}$

(c) $y''$ and $y'''$ for $y = f (x)g(x)$

Show answer

(a) $z' = 120 - t^{2}$, $z'' = -2t$, $z'''= -2$.

(b) $f'(z) = -400z^{-5}$, $f''(z) = 2000z^{-6}$, $f'''(z) = -12000z^{-7}$, $f^{(4)} (z) = 84000z^{-8}$. This last espression evaluated at $z=1$ yields $f^{(4)} (1) = 84000$.

(c) With simplified notation, we obtain $y'= f' g + f g'$. So $y'' = f'' g + f'g' + f'g' + fg''$ by applying the product and sum rule. For the third derivative, we need to once again apply these rules:

$y''' = f''' g + f'' g' + f''g' + f' g'' + f'' g' + f'g'' + f'g'' + fg'''$

$\ \ \ \ \ \ = f'''g + 3f''g' + 3f'g'' + f g'''$.

Question 24

If $u(y)$ denotes an individual’s utility of having income $y$, then $R = −y \cdot u''(y)/u'(y)$ is the coefficient of relative risk aversion. Compute $R$ for the following utility functions, where $A_1$ is a positive constant and we assume that $y > 0$:

(a) $u(y) = A_1 y$

(b) $u(y) = \sqrt{y}$

Show answer

(a) Note that $u'(y) = A_{1}$ and $u''(y) = 0$. Thus, $R=0$.

(b) $u'(y) = (1/2)y^{-1/2}$ and $u''(y) = -(1/4)y^{-3/2}$. This yields $R = \frac{y(1/4)y^{-3/2}}{(1/2)y^{-1/2}} = \frac{0.25}{0.5} = \frac{1}{2}.$