30 Homework: Chain Rule

  1. Watch this video from Khan Academy:
    Chain Rule Definition and example

  2. Take the derivative of the following functions, each of which involves the chain rule.
    1. [latex]a(x) = (x^2 + 5)^{20}[/latex]
      [latex]40 x (x^2 + 5)^{19}[/latex]

      ans
    2. [latex]b(x) = e^{x^2}[/latex]
      [latex]2x e^{x^2}[/latex]

      ans
    3. [latex]c(x) = (kx + r)^n[/latex] for constants [latex]k[/latex], [latex]r[/latex], [latex]n[/latex].
      [latex]k n (kx + r)^{n-1}[/latex]

      ans
    4. [latex]d(x) = (\ln(x))^3 + \ln(x^3)[/latex]
      [latex]\frac{3 (\ln(x))^2}{x} + \frac{3}{x}[/latex]

      ans
    5. [latex]e(x) = \sin(\cos(x))[/latex]
      [latex]-\sin(x) \cos(\cos(x))[/latex]

      ans
    6. [latex]f(x) = e^{\sin(x) + \cos(x)}[/latex]
      [latex](\cos(x) - \sin(x)) e^{\sin(x) + \cos(x)}[/latex]

      ans
    7. [latex]g(x) = \sqrt{3x^2 - 5x + 6}[/latex]
      [latex]\frac{6x - 5}{2 \sqrt{3x^2 - 5x + 6}}[/latex]

      ans
    8. [latex]h(x) = e^{-x}[/latex]
      [latex]-e^{-x}[/latex]

      ans
  3. For each problem, try simplifying the logarithm first, then taking the derivative.
    1. [latex]\frac{d}{dx} \ln(x^3)[/latex]
      [latex]\frac{3}{x}[/latex]

      ans
    2. [latex]\frac{d}{dx} \ln(x e^x)[/latex]
      [latex]\frac{1}{x} + 1[/latex]

      ans
  4. Use logarithm rules to explain why [latex]\frac{d}{dx} \ln(e^5 \cdot x) = \frac{d}{dx} \ln(x)[/latex].
    Using logarithm rules, we have that [latex]ln(e^5 \cdot x) = \ln(x) + \ln(e^5) = \ln(x) + 5[/latex]. This has the same derivative as [latex]\ln(x)[/latex] since we are just adding a constant.

    ans
  5. Recall that [latex]\ln(x)[/latex] and [latex]e^x[/latex] are inverse functions. This means that [latex]\ln(e^x) = x[/latex], and [latex]e^{\ln(x)} = x[/latex] (that is, the [latex]e[/latex] and the [latex]\ln[/latex] cancel out if you do one right after the other). This fact allows us to compute [latex]\frac{d}{dx} 2^x[/latex].
    1. Simplify [latex]e^{\ln(2)}[/latex]
      [latex]=2[/latex]

      ans
    2. Simplify [latex]\ln(e^2)[/latex].
      [latex]=2[/latex]

      ans
    3. Simplify [latex]e^{\ln(2) + x}[/latex]
      [latex]2e^x[/latex]

      ans
    4. Simplify [latex](e^{\ln(2) x})[/latex]
      [latex]2^x[/latex]

      ans

    5. Use part (d) to compute [latex]\frac{d}{dx} 2^x[/latex].
      [latex]\ln(2) 2^x[/latex]

      ans

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Informal Calculus Copyright © by Tyler Seacrest is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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