Week 6 Worksheet

  1. Differentiate the following functions:

    1. \(w(t) = \sin(7t)\)

      Solution

      \(w'(t) = 7\cos(7t)\)

    2. \(f(x) = (3x^4 - 2x^2 + 8)^4\)

      Solution

      \(f'(x) = 4(3x^4 - 2x^2 + 8)^3(12x^2 - 4x)\)

    3. \(g(t) = -3\sqrt{7t^3 - 1}\)

      Solution

      \(g'(t) = \frac{-3}{2\sqrt{7t^3 - 1}}\cdot 21 t^2\)

    4. \(v(x) = \frac{x^2 - 4x}{x + 1}\)

      Solution

      \(v'(x) = \frac{(2x - 4)(x + 1) - (x^2 - 4x)}{(x + 1)^2} = \frac{x^2 + 2x - 4}{(x + 1)^2}\)

    5. \(k(u) = \frac{\sqrt{u} + u}{u^2}\)

      Solution

      This one is easier if we simplify first: \(k(u) = u^{-3/2} + u^{-1}\). Then by the power rule, \(k'(u) = -\frac{3}{2} u^{-5/2} - u^{-2}\).

    6. \(l(q) = e^{q^2 + 7q - 1}\)

      Solution

      \(l'(q) = (2q + 7) e^{q^2 + 7q - 1}\)

  2. Suppose that \(f(2) = 4\), \(f'(2) = 5\), \(g(2) = 7\), and \(g'(2) = -1\). What is \((fg)'(2)\)?

    Solution

    Use the product rule: \((fg)'(2) = f'(2)g(2) + f(2)g'(2) = 5\cdot 7 + 4 \cdot (-1) = 31\).

  3. Values of the functions \(f\) and \(g\), along with their derivatives, are given in the following table:

    \(x\) \(1\) \(2\) \(3\) \(4\)
    \(f(x)\) \(2\) \(4\) \(1\) \(3\)
    \(f'(x)\) \(-6\) \(-7\) \(-8\) \(-9\)
    \(g(x)\) \(2\) \(3\) \(4\) \(1\)
    \(g'(x)\) \(2/7\) \(3/7\) \(4/7\) \(5/7\)

    Compute the following derivatives at \(x=1\):

    1. \(\frac{d}{dx}\left[f(x)g(x)\right]\)

      Solution

      \(\frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)\). Plugging in \(x = 1\) gives \(f'(1)g(1) + f(1)g'(1) = (-6)(2) + (2)(2/7) = \frac{-80}{7}\).

    2. \(\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]\)

      Solution

      \(\frac{d}{dx} \left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}\). Plugging in \(x = 1\) gives

      \[ \frac{f'(1)g(1) - f(1)g'(1)}{g(1)^2} = \frac{(-6)(2) - (2)(2/7)}{(2)^2} = -\frac{22}{7} \]

    3. \(\frac{d}{dx}\left[f(g(x))\right]\)

      Solution

      \(\frac{d}{dx} [f(g(x))] = f'(g(x))g'(x)\). Plugging in \(x = 1\) gives

      \[ f'(g(1))g'(1) = f'(2)\frac{2}{7} = (-7)\left(\frac{2}{7}\right) = -2 \]

    4. \(\frac{d}{dx}\left[g(f(x))\right]\)

      Solution

      \(\frac{d}{dx} [g(f(x))] = g'(f(x))f'(x)\). Plugging in \(x = 1\) gives

      \[ g'(f(1))f'(1) = g'(2)\cdot (-6) = \left(\frac{-3}{7}\right) (-6) = \frac{18}{7} \]

  4. Gross Domestic Product, or GDP, is a statistic that measures the total amount of goods and services produced in a country in a year. Often we want to consider GDP per capita, meaning GDP divided by the total population. As of September 2018, the Bureau of Economic Analysis estimates that the GDP in the United States is $18.57 trillion, and is increasing at a rate of $814 billion dollars per year. According to the CIA World Factbook, the US population is currently 326 million, and increasing at a rate of 2.66 million people per year. What is the rate of change of GDP per capita?

    Solution

    We have \((\text{GDP per capita})(t) = \frac{\operatorname{GDP}(t)}{\operatorname{population}(t)}\), where everything is a function of time. By the quotient rule,

    \[ (\text{GDP per capita})'(t) = \frac{\operatorname{GDP}'(t)\operatorname{population}(t) - \operatorname{GDP}(t)(\operatorname{population})'(t)}{(\operatorname{population}(t))^2} \]

    Plugging in for \(t = (\text{current year})\), we get

    \begin{align*} (\text{GDP per capita})'(t) &= \frac{(\$814 \times 10^9 /\text{year})(326 \times 10^6 \text{ people}) - (\$18.57 \times 10^{12})(2.66 \times 10^6 \text{ people/year})} {(326 \times 10^6 \text{ people})^2} \\ &\approx \$2032.14 / (\text{people} \cdot \text{years}) \end{align*}

    This means that GDP per capita increases by about \$2032.14 per year.

  5. Suppose that \(f\) and \(g\) are two positive, increasing functions (i.e. \(f(x)\), \(f'(x)\), \(g(x)\), \(g'(x)\) are all positive numbers). Let \(h(x) = \frac{f(x)}{g(x)}\). Your friend Gottfried claims that as long as \(f(x)\) is increasing faster than \(g(x)\) (i.e. \(f'(x) > g'(x)\)), we know that \(h(x)\) should be increasing. Is Gottfried right or wrong? If he is right, give a convincing argument. If he is wrong, give a counterexample.

    Solution

    Gottfried is wrong. As a counterexample, take \(f(x) = 2x + 3\) and \(g(x) = x + 1\). Then \(f'(x) = 3\) and \(g'(x) = 1\), so \(f'(x) > g'(x)\). By the quotient rule, the derivative of \(f/g\) at \(x = 0\) is \(\frac{2 \cdot 1 - 1 \cdot 3}{1^2} = -1\), so \(f/g\) is decreasing.

  6. If \(f\) is a positive-valued, differentiable function, I claim that \(\frac{d}{dx} f(x) = f(x) \cdot \frac{d}{dx} \ln [f(x)]\). Give a convincing argument for this claim.

    For some functions, differentiating \(\ln[f(x)]\) is easier than differentiating \(f(x)\) directly. Using this rule, compute the derivative of \(f(x) = x^x\).

    Solution

    Suppose \(f\) is a differentiable function. Then by the chain rule we have

    \[ \frac{d}{dx} [\ln f(x)] = \frac{1}{f(x)} \cdot \frac{d}{dx} f(x) \]

    Re-arranging this gives

    \[ \frac{d}{dx} f(x) = f(x) \cdot \frac{d}{dx} [\ln f(x)] \]

    which is what we wanted to prove.

    To apply this to the function \(f(x) = x^x\), write \(\ln f(x) = \ln (x^x) = x \ln x\). By the product rule,

    \[ \frac{d}{dx} x \ln x = \ln x + \frac{x}{x} = \ln x + 1 \]

    Using the formula we just proved, this means

    \[ f'(x) = x^x(\ln x + 1) \]