3.12: Homework- Chain Rule
- Page ID
- 88652
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- 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.
- \(a(x) = (x^2 + 5)^{20}\)
\(40 x (x^2 + 5)^{19}\)ans
- \(b(x) = e^{x^2}\)
\(2x e^{x^2}\)ans
- \(c(x) = (kx + r)^n\) for constants \(k\), \(r\), \(n\).
\(k n (kx + r)^{n-1}\)ans
- \(d(x) = (\ln(x))^3 + \ln(x^3)\)
\(\frac{3 (\ln(x))^2}{x} + \frac{3}{x}\)ans
- \(e(x) = \sin(\cos(x))\)
\(-\sin(x) \cos(\cos(x))\)ans
- \(f(x) = e^{\sin(x) + \cos(x)}\)
\((\cos(x) - \sin(x)) e^{\sin(x) + \cos(x)}\)ans
- \(g(x) = \sqrt{3x^2 - 5x + 6}\)
\(\frac{6x - 5}{2 \sqrt{3x^2 - 5x + 6}}\)ans
- \(h(x) = e^{-x}\)
\(-e^{-x}\)ans
- \(a(x) = (x^2 + 5)^{20}\)
- For each problem, try simplifying the logarithm first, then taking the derivative.
- \(\frac{d}{dx} \ln(x^3)\)
\(\frac{3}{x}\)ans
- \(\frac{d}{dx} \ln(x e^x)\)
\(\frac{1}{x} + 1\)ans
- \(\frac{d}{dx} \ln(x^3)\)
- 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
- 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 \(\frac{d}{dx} 2^x\).
- Simplify \(e^{\ln(2)}\)
\(=2\)ans
- Simplify \(\ln(e^2)\).
\(=2\)ans
- Simplify \(e^{\ln(2) + x}\)
\(2e^x\)ans
- Simplify \((e^{\ln(2) x})\)
\(2^x\)ans
- Use part (d) to compute \(\frac{d}{dx} 2^x\).
\(\ln(2) 2^x\)ans
- Simplify \(e^{\ln(2)}\)