Fractions and Exponents Practice: Upper Intermediate

Differentiating fractions and powers

Find \( f'(x) \). Rewrite as powers first, differentiate with \( (x^n)' = n x^{n-1} \), then rewrite the answer without negative exponents.

1. 1.
   \( \displaystyle f(x) = \frac{1}{x} \)
   Solution

   \( \displaystyle f(x) = x^{-1} \)
   \( \displaystyle f'(x) = -x^{-2} \)
   \( \displaystyle = -\frac{1}{x^2} \)
2. 2.
   \( \displaystyle f(x) = \frac{3}{x^2} \)
   Solution

   \( \displaystyle f(x) = 3x^{-2} \)
   \( \displaystyle f'(x) = 3 \cdot (-2)\, x^{-3} \)
   \( \displaystyle = -\frac{6}{x^3} \)

   The exponent goes down by one: \( -2 - 1 = -3 \), not \( -1 \).

3. 3.
   \( \displaystyle f(x) = \frac{5}{x^4} \)
   Solution

   \( \displaystyle f(x) = 5x^{-4} \)
   \( \displaystyle f'(x) = -20x^{-5} \)
   \( \displaystyle = -\frac{20}{x^5} \)
4. 4.
   \( \displaystyle f(x) = \frac{1}{2x^3} \)
   Solution

   \( \displaystyle f(x) = \frac12 x^{-3} \)
   \( \displaystyle f'(x) = \frac12 \cdot (-3)\, x^{-4} \)
   \( \displaystyle = -\frac{3}{2x^4} \)
5. 5.
   \( \displaystyle f(x) = \sqrt{x} \)
   Solution

   \( \displaystyle f(x) = x^{1/2} \)
   \( \displaystyle f'(x) = \frac12 x^{-1/2} \)
   \( \displaystyle = \frac{1}{2\sqrt x} \)

   Exponent minus one: \( \frac12 - 1 = -\frac12 \).

6. 6.
   \( \displaystyle f(x) = 6\sqrt[3]{x} \)
   Solution

   \( \displaystyle f(x) = 6x^{1/3} \)
   \( \displaystyle f'(x) = 6 \cdot \frac13\, x^{-2/3} \)
   \( \displaystyle = \frac{2}{x^{2/3}} \)
7. 7.
   \( \displaystyle f(x) = \frac{4}{\sqrt{x}} \)
   Solution

   \( \displaystyle f(x) = 4x^{-1/2} \)
   \( \displaystyle f'(x) = 4 \cdot \left(-\frac12\right) x^{-3/2} \)
   \( \displaystyle = -\frac{2}{x^{3/2}} \)

   \( -\frac12 - 1 = -\frac32 \). Subtracting 1 from a negative fraction makes it more negative.

8. 8.
   \( \displaystyle f(x) = x\sqrt{x} \)
   Solution

   \( \displaystyle f(x) = x^{3/2} \)
   \( \displaystyle f'(x) = \frac32 x^{1/2} \)
   \( \displaystyle = \frac{3\sqrt x}{2} \)

   Combining into one power first avoids the product rule.

9. 9.
   \( \displaystyle f(x) = \frac{\sqrt{x}}{x^2} \)
   Solution

   \( \displaystyle f(x) = x^{\frac12 - 2} \)
   \( \displaystyle = x^{-3/2} \)
   \( \displaystyle f'(x) = -\frac32 x^{-5/2} \)
   \( \displaystyle = -\frac{3}{2x^{5/2}} \)
10. 10.
    \( \displaystyle f(x) = \left(x^{1/3}\right)^2 \cdot x \)
    Solution

    \( \displaystyle f(x) = x^{2/3} \cdot x^{1} \)
    \( \displaystyle = x^{5/3} \)
    \( \displaystyle f'(x) = \frac53 x^{2/3} \)
11. 11.
    \( \displaystyle f(x) = \frac{x}{2} + \frac{2}{x} \)
    Solution

    \( \displaystyle f(x) = \frac12 x + 2x^{-1} \)
    \( \displaystyle f'(x) = \frac12 - 2x^{-2} \)
    \( \displaystyle = \frac12 - \frac{2}{x^2} \)
12. 12.
    \( \displaystyle f(x) = \frac{x^2 + 1}{x} \)
    Solution

    \( \displaystyle f(x) = \frac{x^2}{x} + \frac{1}{x} \)
    \( \displaystyle = x + x^{-1} \)
    \( \displaystyle f'(x) = 1 - x^{-2} \)
    \( \displaystyle = 1 - \frac{1}{x^2} \)

    Split the fraction term by term (the denominator is a single term), then each piece is a power.

13. 13.
    \( \displaystyle f(x) = \frac{x^3 - 2x}{x^2} \)
    Solution

    \( \displaystyle f(x) = x - 2x^{-1} \)
    \( \displaystyle f'(x) = 1 + 2x^{-2} \)
    \( \displaystyle = 1 + \frac{2}{x^2} \)

    Minus times minus: the derivative of \( -2x^{-1} \) is \( (-2)(-1)x^{-2} = +2x^{-2} \).

14. 14.
    \( \displaystyle f(x) = \frac{2x + 3}{\sqrt{x}} \)
    Solution

    \( \displaystyle f(x) = 2x^{1/2} + 3x^{-1/2} \)
    \( \displaystyle f'(x) = x^{-1/2} - \frac32 x^{-3/2} \)
    \( \displaystyle = \frac{2x - 3}{2x^{3/2}} \)

    To write the answer as one fraction, put both terms over \( 2x^{3/2} \): \( x^{-1/2} = \frac{2x}{2x^{3/2}} \).

15. 15.
    \( \displaystyle f(x) = \frac{1}{1 + x} \)
    Solution

    \( \displaystyle f(x) = (1 + x)^{-1} \)
    \( \displaystyle f'(x) = -(1 + x)^{-2} \cdot 1 \)
    \( \displaystyle = -\frac{1}{(1 + x)^2} \)

    When the bottom is a sum you cannot split the fraction. Treat the bracket as one block and use the chain rule (its inside has derivative 1 here).

16. 16.
    \( \displaystyle f(x) = \frac{3}{(2x + 1)^2} \)
    Solution

    \( \displaystyle f(x) = 3(2x + 1)^{-2} \)
    \( \displaystyle f'(x) = 3 \cdot (-2)(2x + 1)^{-3} \cdot 2 \)
    \( \displaystyle = -\frac{12}{(2x + 1)^3} \)

    The final \( \cdot 2 \) is the derivative of the inside, \( 2x + 1 \).

17. 17.
    \( \displaystyle f(x) = \sqrt{1 + 2x} \)
    Solution

    \( \displaystyle f(x) = (1 + 2x)^{1/2} \)
    \( \displaystyle f'(x) = \frac12 (1 + 2x)^{-1/2} \cdot 2 \)
    \( \displaystyle = \frac{1}{\sqrt{1 + 2x}} \)
18. 18.
    \( \displaystyle f(x) = \frac{x}{x + 1} \)
    Solution

    \( \displaystyle f'(x) = \frac{1 \cdot (x + 1) - x \cdot 1}{(x + 1)^2} \)
    \( \displaystyle = \frac{1}{(x + 1)^2} \)

    Variable on top and on the bottom: quotient rule, \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).

19. 19.
    \( \displaystyle f(x) = \frac{x^2}{x + 1} \)
    Solution

    \( \displaystyle f'(x) = \frac{2x(x + 1) - x^2 \cdot 1}{(x + 1)^2} \)
    \( \displaystyle = \frac{x^2 + 2x}{(x + 1)^2} \)

    Expand the top and collect: \( 2x^2 + 2x - x^2 = x^2 + 2x \). Leave the bottom squared; do not expand it.

20. 20.
    \( \displaystyle f(x) = \frac{x^2 + 3}{x - 1} \)
    Solution

    \( \displaystyle f'(x) = \frac{2x(x - 1) - (x^2 + 3)}{(x - 1)^2} \)
    \( \displaystyle = \frac{x^2 - 2x - 3}{(x - 1)^2} \)

    Bracket \( (x^2 + 3) \) before subtracting it: the minus applies to the 3 as well.

Fractions and exponents in economics

The same algebra inside the problems you meet in the course. Every step is shown.

1. 21.
   For \( U(x, y) = x^{1/2} y^{1/2} \), compute the marginal utilities, simplify the marginal rate of substitution \( MRS = U_x / U_y \), and evaluate it at \( (x, y) = (4, 9) \).
   Solution

   \[ U_x = \tfrac12 x^{-1/2} y^{1/2}, \qquad U_y = \tfrac12 x^{1/2} y^{-1/2} \]

   Divide, grouping numbers with numbers and each letter with itself:

   \[ MRS = \frac{\tfrac12}{\tfrac12} \cdot \frac{x^{-1/2}}{x^{1/2}} \cdot \frac{y^{1/2}}{y^{-1/2}} = 1 \cdot x^{-\frac12 - \frac12} \cdot y^{\frac12 + \frac12} = x^{-1} y = \frac{y}{x} \]

   At \( (4, 9) \): \( MRS = \frac94 \). Subtracting a negative exponent is where most mistakes happen: \( \frac12 - \left(-\frac12\right) = 1 \).

2. 22.
   For \( U(x, y) = x^{1/4} y^{3/4} \), show that \( MRS = \frac{y}{3x} \). A consumer with income 40 faces \( p_x = 3 \) and \( p_y = 1 \). Set the MRS equal to the price ratio and use the budget to find the optimal bundle.
   Solution

   \[ MRS = \frac{\tfrac14 x^{-3/4} y^{3/4}}{\tfrac34 x^{1/4} y^{-1/4}} = \frac{1/4}{3/4} \cdot x^{-\frac34 - \frac14} \cdot y^{\frac34 + \frac14} = \frac13\, x^{-1} y = \frac{y}{3x} \]

   The fraction of fractions: \( \frac{1/4}{3/4} = \frac14 \cdot \frac43 = \frac13 \). Now the tangency condition:

   \[ \frac{y}{3x} = \frac{p_x}{p_y} = \frac31 \quad\Longrightarrow\quad y = 9x \]

   Into the budget \( 3x + y = 40 \): \( 3x + 9x = 40 \), so \( x = \frac{40}{12} = \frac{10}{3} \) and \( y = 9 \cdot \frac{10}{3} = 30 \). Check: spending on \( x \) is \( 3 \cdot \frac{10}{3} = 10 \), a quarter of income, the exponent on \( x \).

3. 23.
   A consumer has \( U(x, y) = x^{1/3} y^{2/3} \), income 60 and prices \( p_x = 2 \), \( p_y = 4 \). Find the optimal bundle.
   Solution

   The MRS was simplified on the Intermediate sheet: \( MRS = \frac{y}{2x} \). Tangency:

   \[ \frac{y}{2x} = \frac{2}{4} = \frac12 \quad\Longrightarrow\quad y = \frac12 \cdot 2x = x \]

   Budget: \( 2x + 4y = 60 \) with \( y = x \) gives \( 6x = 60 \), so \( x = 10 \) and \( y = 10 \). Spending: \( 2 \cdot 10 = 20 \) on \( x \), one third of 60, and \( 40 \) on \( y \), two thirds. The exponents are the budget shares.

4. 24.
   The general case. Show that \( U = x^{a} y^{1-a} \) with budget \( p_x x + p_y y = m \) gives the demands \( x^* = \frac{a\,m}{p_x} \) and \( y^* = \frac{(1-a)\,m}{p_y} \).
   Solution

   First the MRS, subtracting exponents that contain letters:

   \[ MRS = \frac{a\, x^{a-1} y^{1-a}}{(1-a)\, x^{a} y^{-a}} = \frac{a}{1-a}\, x^{(a-1) - a}\, y^{(1-a) + a} = \frac{a}{1-a} \cdot \frac{y}{x} \]

   Tangency, \( MRS = \frac{p_x}{p_y} \), then multiply both sides by \( x\, p_y \) (cross-multiplying):

   \[ \frac{a\,y}{(1-a)\,x} = \frac{p_x}{p_y} \quad\Longrightarrow\quad a\, p_y y = (1-a)\, p_x x \quad\Longrightarrow\quad p_y y = \frac{1-a}{a}\, p_x x \]

   Into the budget:

   \[ p_x x + \frac{1-a}{a}\, p_x x = m \quad\Longrightarrow\quad p_x x\left(\frac{a}{a} + \frac{1-a}{a}\right) = m \quad\Longrightarrow\quad p_x x \cdot \frac{1}{a} = m \]

   so \( x^* = \frac{a\,m}{p_x} \), and then \( p_y y = m - a\,m = (1-a)\,m \) gives \( y^* = \frac{(1-a)\,m}{p_y} \). The key fraction step: \( 1 + \frac{1-a}{a} = \frac{a + 1 - a}{a} = \frac1a \).

5. 25.
   A firm has \( Q(K, L) = K^{1/2} L^{1/2} \) and \( K = 16 \) machines. Find the marginal product of labour, and the amount of labour at which it equals 1.
   Solution

   \[ MP_L = \frac{\partial Q}{\partial L} = \tfrac12 K^{1/2} L^{-1/2} = \frac12 \cdot 16^{1/2} \cdot L^{-1/2} = \frac12 \cdot 4 \cdot L^{-1/2} = \frac{2}{\sqrt L} \]

   Set it equal to 1:

   \[ \frac{2}{\sqrt L} = 1 \quad\Longrightarrow\quad \sqrt L = 2 \quad\Longrightarrow\quad L = 4 \]

   Squaring undoes the square root. The marginal product falls as \( L \) grows, because \( L \) sits under a root in the denominator.

6. 26.
   A firm's profit from hiring \( L \) workers is \( \pi(L) = 20\sqrt{L} - 5L \). Find the profit-maximising \( L \), check it is a maximum, and compute the profit.
   Solution

   Rewrite the root as a power, then differentiate:

   \[ \pi(L) = 20 L^{1/2} - 5L, \qquad \pi'(L) = 10 L^{-1/2} - 5 = \frac{10}{\sqrt L} - 5 \] \[ \frac{10}{\sqrt L} = 5 \quad\Longrightarrow\quad \sqrt L = 2 \quad\Longrightarrow\quad L^* = 4 \]

   Second derivative: \( \pi''(L) = 10 \cdot \left(-\tfrac12\right) L^{-3/2} = -\frac{5}{L^{3/2}} < 0 \), a maximum. Profit: \( \pi(4) = 20 \cdot 2 - 20 = 20 \).

7. 27.
   A firm has cost \( C(q) = 50 + 2q^2 \). Write average cost \( AC(q) = C(q)/q \) as a sum of powers of \( q \), find the output that minimises it, and the minimum average cost.
   Solution

   Split the fraction term by term (the denominator \( q \) is a single term):

   \[ AC(q) = \frac{50}{q} + \frac{2q^2}{q} = 50 q^{-1} + 2q \] \[ AC'(q) = -50 q^{-2} + 2 = -\frac{50}{q^2} + 2 = 0 \quad\Longrightarrow\quad q^2 = 25 \quad\Longrightarrow\quad q^* = 5 \]

   Minimum average cost: \( AC(5) = \frac{50}{5} + 10 = 20 \). The fixed cost 50 is spread over more units as \( q \) grows (the \( \frac{50}{q} \) term falls), while the \( 2q \) term rises: the minimum balances the two.

8. 28.
   Demand is \( Q(p) = 100\, p^{-3/2} \). Show that the price elasticity \( \varepsilon = Q'(p) \cdot \frac{p}{Q} \) is the same at every price.
   Solution

   \[ Q'(p) = 100 \cdot \left(-\tfrac32\right) p^{-5/2} = -150\, p^{-5/2} \] \[ \varepsilon = -150\, p^{-5/2} \cdot \frac{p}{100\, p^{-3/2}} = -\frac{150}{100} \cdot p^{-\frac52 + 1 + \frac32} = -\frac32 \cdot p^{0} = -\frac32 \]

   Collect all the powers of \( p \) into one exponent: \( -\frac52 + 1 + \frac32 = 0 \). Dividing by \( p^{-3/2} \) is multiplying by \( p^{3/2} \). The elasticity equals the exponent: that is why this is called constant-elasticity demand.

9. 29.
   A monopolist faces inverse demand \( p(Q) = \frac{12}{\sqrt{Q}} \) and has marginal cost 1. Find revenue as a power of \( Q \), marginal revenue, and the profit-maximising quantity and price.
   Solution

   \[ R(Q) = p(Q)\cdot Q = \frac{12}{Q^{1/2}} \cdot Q = 12\, Q^{1 - \frac12} = 12\, Q^{1/2} \] \[ MR(Q) = 6\, Q^{-1/2} = \frac{6}{\sqrt Q} \]

   Set \( MR = MC \): \( \frac{6}{\sqrt Q} = 1 \), so \( \sqrt Q = 6 \) and \( Q^* = 36 \). Price: \( p = \frac{12}{\sqrt{36}} = \frac{12}{6} = 2 \).

10. 30.
    A bond pays 100 in two years and costs 81 today. Find the interest rate \( r \) that solves \( \frac{100}{(1 + r)^2} = 81 \).
    Solution

    \[ \frac{100}{(1 + r)^2} = 81 \quad\Longrightarrow\quad (1 + r)^2 = \frac{100}{81} \quad\Longrightarrow\quad 1 + r = \left(\frac{100}{81}\right)^{1/2} = \frac{10}{9} \]

    so \( r = \frac{10}{9} - 1 = \frac19 \approx 11.1\% \). Swapping \( (1 + r)^2 \) and 81 is cross-multiplying; the square root of a fraction is the root of the top over the root of the bottom.

11. 31.
    Problem Set 1 with numbers. An agent's preferred tax rate is \( \tau_i^* = \frac12\left(1 - \frac{y_i}{\bar y}\right) \). With average income \( \bar y = \frac12 \), compute \( \tau_i^* \) for an agent with income \( y_i = \frac15 \), and for the median voter if median income is \( \frac25 \).
    Solution

    The fraction inside the bracket first, dividing by a fraction by flipping it:

    \[ \frac{y_i}{\bar y} = \frac{1/5}{1/2} = \frac15 \cdot 2 = \frac25, \qquad \tau_i^* = \frac12\left(1 - \frac25\right) = \frac12 \cdot \frac35 = \frac{3}{10} \]

    Median voter: \( \frac{2/5}{1/2} = \frac45 \), so \( \tau^* = \frac12 \cdot \frac15 = \frac{1}{10} \). Poorer agents want higher taxes, and the majority chooses a 10% tax because the median is below the mean.