3.12: Homework- Chain Rule - Mathematics LibreTexts

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  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. \(a(x) = (x^2 + 5)^{20}\) \(40 x (x^2 + 5)^{19}\) ans
    2. \(b(x) = e^{x^2}\) \(2x e^{x^2}\) ans
    3. \(c(x) = (kx + r)^n\) for constants \(k\), \(r\), \(n\). \(k n (kx + r)^{n-1}\) ans
    4. \(d(x) = (\ln(x))^3 + \ln(x^3)\) \(\frac{3 (\ln(x))^2}{x} + \frac{3}{x}\) ans
    5. \(e(x) = \sin(\cos(x))\) \(-\sin(x) \cos(\cos(x))\) ans
    6. \(f(x) = e^{\sin(x) + \cos(x)}\) \((\cos(x) - \sin(x)) e^{\sin(x) + \cos(x)}\) ans
    7. \(g(x) = \sqrt{3x^2 - 5x + 6}\) \(\frac{6x - 5}{2 \sqrt{3x^2 - 5x + 6}}\) ans
    8. \(h(x) = e^{-x}\) \(-e^{-x}\) ans
  3. For each problem, try simplifying the logarithm first, then taking the derivative.
    1. \(\frac{d}{dx} \ln(x^3)\) \(\frac{3}{x}\) ans
    2. \(\frac{d}{dx} \ln(x e^x)\) \(\frac{1}{x} + 1\) ans
  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. ans
  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 \(e^{\ln(2)}\) \(=2\) ans
    2. Simplify \(\ln(e^2)\). \(=2\) ans
    3. Simplify \(e^{\ln(2) + x}\) \(2e^x\) ans
    4. Simplify \((e^{\ln(2) x})\) \(2^x\) ans
    5. Use part (d) to compute \(\frac{d}{dx} 2^x\). \(\ln(2) 2^x\) ans

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