3.12: Homework- Chain Rule
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Watch this video from Khan Academy:
Chain Rule Definition and example -
Take the derivative of the following functions, each of which involves the chain rule.
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\(a(x) = (x^2 + 5)^{20}\)
\(40 x (x^2 + 5)^{19}\)ans
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\(b(x) = e^{x^2}\)
\(2x e^{x^2}\)ans
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\(c(x) = (kx + r)^n\) for constants \(k\), \(r\), \(n\).
\(k n (kx + r)^{n-1}\)ans
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\(d(x) = (\ln(x))^3 + \ln(x^3)\)
\(\frac{3 (\ln(x))^2}{x} + \frac{3}{x}\)ans
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\(e(x) = \sin(\cos(x))\)
\(-\sin(x) \cos(\cos(x))\)ans
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\(f(x) = e^{\sin(x) + \cos(x)}\)
\((\cos(x) - \sin(x)) e^{\sin(x) + \cos(x)}\)ans
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\(g(x) = \sqrt{3x^2 - 5x + 6}\)
\(\frac{6x - 5}{2 \sqrt{3x^2 - 5x + 6}}\)ans
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\(h(x) = e^{-x}\)
\(-e^{-x}\)ans
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\(a(x) = (x^2 + 5)^{20}\)
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For each problem, try simplifying the logarithm first, then taking the derivative.
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\(\frac{d}{dx} \ln(x^3)\)
\(\frac{3}{x}\)ans
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\(\frac{d}{dx} \ln(x e^x)\)
\(\frac{1}{x} + 1\)ans
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\(\frac{d}{dx} \ln(x^3)\)
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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
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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 \(\frac{d}{dx} 2^x\).
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Simplify \(e^{\ln(2)}\)
\(=2\)ans
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Simplify \(\ln(e^2)\).
\(=2\)ans
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Simplify \(e^{\ln(2) + x}\)
\(2e^x\)ans
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Simplify \((e^{\ln(2) x})\)
\(2^x\)ans
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Use part (d) to compute \(\frac{d}{dx} 2^x\).
\(\ln(2) 2^x\)ans
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Simplify \(e^{\ln(2)}\)