3.13: Multirule Derivatives
- Page ID
- 88653
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Okay, let’s talk about \(\frac{d}{dx} \ e^{x^2 + x} \sin(x)\). If you’re thinking this looks like a product rule, but it also looks like a chain rule, you’re right. To compute this derivative, we need to do the chain rule and the product rule. This is because it is a multirule problem. Let’s do this example
The way I like to break this down is to consider a little rule and a big rule. In this case, the little rule is the chain rule problem \(\frac{d}{dx} \ e^{x^2 + x}\). If we do this problem, we see that \(f = e^x\), \(f' = e^x\), \(g = x^2 + x\) and \(g' = 2x + 1\). So we have
\[\begin{equation*} \frac{d}{dx} \ e^{x^2 + x} = {\color{red} e^{x^2 + x} (2x + 1)}. \end{equation*}\]
Now we are ready to do the big rule, which is the product rule. At this point we go back to the original problem \(\frac{d}{dx} \ e^{x^2 + x} \sin(x)\). For this product rule, we see \(f = e^{x^2 + x}\), \(g = \sin(x)\), \(g' = \cos(x)\). What is \({\color{red} f'}\)? Why, that’s what we just computed in the equation above! So \({\color{red} f' = e^{x^2 + x}(2x+1)}\). Putting this all together with the product rule \(f g' + g f'\), we have
\[\begin{align*} \frac{d}{dx} \ e^{x^2 + x} \sin(x) & = f g' + g {\color{red} f'} \\ & = \boxed{e^{x^2 + x} \cos(x) + \sin(x) {\color{red} e^{x^2 + x} (2x + 1)}} . \end{align*}\]
little chain rule: \(\frac{d}{dx} \sin(x^2 + x)\)
\(\begin{array}{ll} f = \sin(x) & g = x^2 + x \\ f' = \cos(x) & g' = 2x + 1 \end{array}\)
Result: \({\color{red} \cos(x^2 + x) \cdot (2x+1)}\)
Big quotient rule (aka the whole problem):\(\frac{d}{dx} \frac{x}{\sin(x^2 + x)}\)
\(\begin{array}{ll} f = x & g = \sin(x^2 + x) \\ f' = 1 & g' = {\color{red} \cos(x^2 + x) \cdot (2x+1)} \end{array}\)
Result: \(\boxed{\frac{\sin(x^2 + x) \cdot 1 - x \cos(x^2 + x) \cdot (2x+1)}{(\sin(x^2+x))^2}}\)