3.12: 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. \(a(x) = (x^2 + 5)^{20}\)
      \(40 x (x^2 + 5)^{19}\)

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   2. \(b(x) = e^{x^2}\)
      \(2x e^{x^2}\)

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   3. \(c(x) = (kx + r)^n\) for constants \(k\), \(r\), \(n\).
      \(k n (kx + r)^{n-1}\)

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   4. \(d(x) = (\ln(x))^3 + \ln(x^3)\)
      \(\frac{3 (\ln(x))^2}{x} + \frac{3}{x}\)

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   5. \(e(x) = \sin(\cos(x))\)
      \(-\sin(x) \cos(\cos(x))\)

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   6. \(f(x) = e^{\sin(x) + \cos(x)}\)
      \((\cos(x) - \sin(x)) e^{\sin(x) + \cos(x)}\)

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   7. \(g(x) = \sqrt{3x^2 - 5x + 6}\)
      \(\frac{6x - 5}{2 \sqrt{3x^2 - 5x + 6}}\)

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   8. \(h(x) = e^{-x}\)
      \(-e^{-x}\)

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

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   2. \(\frac{d}{dx} \ln(x e^x)\)
      \(\frac{1}{x} + 1\)

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

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

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   2. Simplify \(\ln(e^2)\).
      \(=2\)

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   3. Simplify \(e^{\ln(2) + x}\)
      \(2e^x\)

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   4. Simplify \((e^{\ln(2) x})\)
      \(2^x\)

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

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