Tutorial

Math V46A Chapter 4 Selected Worked-Out Solutions

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Solutions

4.3 #31.

$$\begin{array}{l}m(t)=-6t{\left(5t^4-1\right)}^4\text{This is a product rule situation with an embedded chain rule in the}{v}^{\prime }\text{calculation}\text{.}\\ u=-6t;u^{\prime }=-6\\ v={\left(5t^4-1\right)}^4;v^{\prime }=4{\left(5t^4-1\right)}^3\cdot \frac{d}{dt}\left(5t^4-1\right)=4{\left(5t^4-1\right)}^3\left(20t^3\right)=80t^3{\left(5t^4-1\right)}^3\\ m^{\prime }(t)=u^{\prime }v+v^{\prime }u=-6{\left(5t^4-1\right)}^4+80t^3{\left(5t^4-1\right)}^3\left(-6t\right)\\ m^{\prime }(t)=-6{\left(5t^4-1\right)}^4-480t^4{\left(5t^4-1\right)}^3\\ m^{\prime }(t)=-6{\left(5t^4-1\right)}^3\left[\left(5t^4-1\right)+80t^4\right]\text{(GCF factoring)}\\ m^{\prime }(t)=-6{\left(5t^4-1\right)}^3\left[5t^4-1+80t^4\right]=\underset{¯}{\underset{¯}{-6{\left(5t^4-1\right)}^3\left(85t^4-1\right)}}\end{array}$$

4.3 #35.

$$\begin{array}{l}q(y)=4y^2{\left(y^2+1\right)}^{5/4}\text{This is a product rule situation with an embedded chain rule in the}{v}^{\prime }\text{calculation}\text{.}\\ u=4y^2;u^{\prime }=8y\\ v={\left(y^2+1\right)}^{5/4};v^{\prime }=\frac{5}{4}{\left(y^2+1\right)}^{1/4}\cdot \frac{d}{dy}\left(y^2+1\right)=\frac{5}{4}{\left(y^2+1\right)}^{1/4}\left(2y\right)=\frac{5}{2}y{\left(y^2+1\right)}^{1/4}\\ q^{\prime }(y)=u^{\prime }v+v^{\prime }u=8y{\left(y^2+1\right)}^{5/4}+\frac{5}{2}y{\left(y^2+1\right)}^{1/4}\left(4y^2\right)\\ q^{\prime }(y)=8y{\left(y^2+1\right)}^{5/4}+10y^3{\left(y^2+1\right)}^{1/4}\\ q^{\prime }(y)=2y{\left(y^2+1\right)}^{1/4}\left[4\left(y^2+1\right)+5y^2\right]\text{(GCF factoring)}\\ q^{\prime }(y)=2y{\left(y^2+1\right)}^{1/4}\left[4y^2+4+5y^2\right]=\underset{¯}{\underset{¯}{2y{\left(y^2+1\right)}^{1/4}\left(9y^2+4\right)}}\end{array}$$

4.3 #39.

$$\begin{array}{l}r(t)=\frac{{\left(5t-6\right)}^4}{3t^2+4}\text{This is a quotient rule situation with an embedded chain rule in the}{u}^{\prime }\text{calculation}\text{.}\\ u={\left(5t-6\right)}^4;u^{\prime }=4{\left(5t-6\right)}^3\cdot \frac{d}{dt}\left(5t-6\right)=4{\left(5t-6\right)}^3\left(5\right)=20{\left(5t-6\right)}^3\\ v=\left(3t^2+4\right);v^{\prime }=6t\\ r^{\prime }(t)=\frac{u^{\prime }v-v^{\prime }u}{v^2}=\frac{20{\left(5t-6\right)}^3\left(3t^2+4\right)-6t{\left(5t-6\right)}^4}{{\left(3t^2+4\right)}^2}\\ r^{\prime }(t)=\frac{2{\left(5t-6\right)}^3\left[10\left(3t^2+4\right)-3t\left(5t-6\right)\right]}{{\left(3t^2+4\right)}^2}\text{(GCF factoring)}\\ r^{\prime }(t)=\frac{2{\left(5t-6\right)}^3\left[30t^2+40-15t^2+18t\right]}{{\left(3t^2+4\right)}^2}\text{(watch the signs in the second distributive inside the brackets)}\\ \underset{¯}{\underset{¯}{r^{\prime }(t)=\frac{2{\left(5t-6\right)}^3\left[15t^2+18t+40\right]}{{\left(3t^2+4\right)}^2}}}\end{array}$$

4.3 #57.

$$\begin{array}{l}V(t)=\frac{60,000}{1+0.3t+0.1t^2}\text{The rate at which the value is changing is given by}{V}^{\prime }(t)\text{; use the quotient rule}\text{.}\\ u=60,000;u^{\prime }=0\\ v=\left(1+0.3t+0.1t^2\right);v^{\prime }=\left(0.3+0.2t\right)\\ V^{\prime }(t)=\frac{u^{\prime }v-v^{\prime }u}{v^2}=\frac{0\left(1+0.3t+0.1t^2\right)-\left(0.3+0.2t\right)\left(60,000\right)}{{\left(1+0.3t+0.1t^2\right)}^2}\\ V^{\prime }(t)=\frac{-60,000\left(0.3+0.2t\right)}{{\left(1+0.3t+0.1t^2\right)}^2}=\frac{100\left[-60,000\left(0.3+0.2t\right)\right]}{100{\left(1+0.3t+0.1t^2\right)}^2}\text{(multiply top and bottom by 100 to get rid of decimals)}\\ V^{\prime }(t)=\frac{\left[\left(10\right)\left(-60,000\right)\right]\cdot \left[10\left(0.3+0.2t\right)\right]}{{\left(10\right)}^2{\left(1+0.3t+0.1t^2\right)}^2}\text{(rewrite both 100s in funny ways so they multiply more easily)}\\ V^{\prime }(t)=\frac{-600,000\left(3+2t\right)}{{\left[\left(10\right)\left(1+0.3t+0.1t^2\right)\right]}^2}=\underset{¯}{\underset{¯}{\frac{-600,000\left(3+2t\right)}{{\left(10+3t+t^2\right)}^2}}}\text{(generic derivative; no more decimals!)}\\ \text{Part}(a)\text{:}{V}^{\prime }(2)=\frac{-600,000\left[3+2\left(2\right)\right]}{{\left[10+3\left(2\right)+{\left(2\right)}^2\right]}^2}=\frac{-600,000\left[7\right]}{{\left[10+6+4\right]}^2}=\frac{-4,200,000}{{\left(20\right)}^2}=\frac{-4,200,000}{400}=\underset{¯}{\underset{¯}{-10,500}}\\ \text{Part}(b)\text{:}{V}^{\prime }(4)=\frac{-600,000\left[3+2\left(4\right)\right]}{{\left[10+3\left(4\right)+{\left(4\right)}^2\right]}^2}=\frac{-600,000\left[11\right]}{{\left[10+12+16\right]}^2}=\frac{-6,600,000}{{\left(38\right)}^2}=\frac{-6,600,000}{1444}\approx \underset{¯}{\underset{¯}{-4,570.64}}\end{array}$$

4.4 #15.

$$\begin{array}{l}y={\left(x+3\right)}^2{e}^{4x}\text{This is a product rule situation with embedded chain rules in both the}{u}^{\prime }\text{and}{v}^{\prime }\text{calculations}\text{.}\\ u={\left(x+3\right)}^2;u^{\prime }=2\left(x+3\right)\cdot \frac{d}{dx}\left(x+3\right)=2\left(x+3\right)\left(1\right)=2\left(x+3\right)\\ v=e^{4x};v^{\prime }=e^{4x}\cdot \frac{d}{dx}\left(4x\right)=4e^{4x}\\ y^{\prime }=u^{\prime }v+v^{\prime }u=2\left(x+3\right)e^{4x}+4e^{4x}{\left(x+3\right)}^2\\ y^{\prime }=2\left(x+3\right)e^{4x}\left[1+2\left(x+3\right)\right]\text{(GCF factoring)}\\ y^{\prime }=2\left(x+3\right)\left(1+2x+6\right)e^{4x}=\underset{¯}{\underset{¯}{2\left(x+3\right)\left(2x+7\right)e^{4x}}}\end{array}$$

4.4 #31.

$$\begin{array}{l}y=\frac{t^2{e}^{2t}}{t+e^{3t}}\text{This is a quotient rule situation with an embedded product rule upstairs, plus multiple chain rules}\text{.}\\ \text{To avoid confusion, let's call the numerator}u\text{and the denominator}v\text{; within the numerator, let}p=t^2\text{and}q=e^{2t}.\\ \text{Then, when working on the product embedded within the numerator, the product rule will read}{p}^{\prime }q+q^{\prime }p.\\ u=pq=t^2{e}^{2t};u^{\prime }=p^{\prime }q+q^{\prime }p=\left(2t\right)\left(e^{2t}\right)+\left[e^{2t}\cdot \frac{d}{dt}\left(2t\right)\right]\left(t^2\right)=2t{e}^{2t}+\left[2e^{2t}\right]\left(t^2\right)=2t{e}^{2t}+2t^2{e}^{2t}=2t{e}^{2t}\left(1+t\right)\\ v=t+e^{3t};v^{\prime }=1+e^{3t}\cdot \frac{d}{dt}\left(3t\right)=1+3e^{3t}\\ y^{\prime }=\frac{u^{\prime }v-v^{\prime }u}{v^2}=\frac{\left[2t{e}^{2t}\left(1+t\right)\right]\left(t+e^{3t}\right)-\left(1+3e^{3t}\right)\left(t^2{e}^{2t}\right)}{{\left(t+e^{3t}\right)}^2}\\ y^{\prime }=\frac{2t{e}^{2t}\left[\left(1+t\right)\left(t+e^{3t}\right)\right]-t^2{e}^{2t}\left(1+3e^{3t}\right)}{{\left(t+e^{3t}\right)}^2}\text{(associative law regroups the factors in front of the minus sign upstairs}\dots \text{)}\\ y^{\prime }=\frac{2t{e}^{2t}\left[t+e^{3t}+t^2+t{e}^{3t}\right]-t^2{e}^{2t}\left(1+3e^{3t}\right)}{{\left(t+e^{3t}\right)}^2}\text{(}\dots \text{which makes them easier to FOIL)}\\ y^{\prime }=\frac{2t^2{e}^{2t}+2t{e}^{5t}+2t^3{e}^{2t}+2t^2{e}^{5t}-t^2{e}^{2t}-3t^2{e}^{5t}}{{\left(t+e^{3t}\right)}^2}=\underset{¯}{\underset{¯}{\frac{e^{2t}\left(2t^3+t^2\right)+e^{5t}\left(2t-t^2\right)}{{\left(t+e^{3t}\right)}^2}}}\end{array}$$

4.5 #35.

$$\begin{array}{l}y=\left({\log }_7\sqrt{4x-3}\right)\text{This is a logarithm containing two nested chain rules (square root and then the stuff inside)}\text{.}\\ \text{Ugly! Use the properties of the logarithm to reduce the chain rule layers from two to one}before\text{differentiating}\text{.}\\ y=\left[{\log }_7{\left(4x-3\right)}^{1/2}\right]=\frac{\text{1}}{\text{2}}\left[{\log }_7\left(4x-3\right)\right]\text{(much nicer, and no calculus required}\dots \text{yet)}\\ y^{\prime }=\frac{d}{dx}\left[\frac{1}{2}{\log }_7\left(4x-3\right)\right]=\frac{1}{2}\frac{d}{dx}\left[{\log }_7\left(4x-3\right)\right]\text{(constant rule)}\\ y^{\prime }=\frac{1}{2}\left[\frac{1}{4x-3}\cdot \frac{1}{\ln 7}\right]\cdot \frac{d}{dx}\left(4x-3\right)\text{(log rule inside the brackets, followed by a promised chain rule)}\\ y^{\prime }=\frac{1}{2\ln 7}\left(\frac{1}{4x-3}\right)\cdot \left(4\right)=\frac{4}{2\ln 7}\left(\frac{1}{4x-3}\right)=\underset{¯}{\underset{¯}{\frac{2}{\left(\ln 7\right)\left(4x-3\right)}}}\\ \text{If you insist on keeping the square root (forgot your logs?), at least change it to a power (alternative approach):}\\ y^{\prime }=\frac{d}{dx}\left[{\log }_7{\left(4x-3\right)}^{1/2}\right]=\left[\frac{1}{{\left(4x-3\right)}^{1/2}}\cdot \frac{1}{\ln 7}\right]\cdot \frac{d}{dx}{\left(4x-3\right)}^{1/2}\text{(first layer of chain rule promised on the right)}\\ y^{\prime }=\left[\frac{1}{{\left(4x-3\right)}^{1/2}}\cdot \frac{1}{\ln 7}\right]\cdot \frac{1}{2}{\left(4x-3\right)}^{-1/2}\cdot \frac{d}{dx}\left(4x-3\right)\text{(second layer of chain rule)}\\ y^{\prime }=\left[\frac{1}{{\left(4x-3\right)}^{1/2}}\cdot \frac{1}{\ln 7}\right]\cdot \frac{1}{2}\cdot \frac{1}{{\left(4x-3\right)}^{1/2}}\cdot \left(4\right)\text{(got rid of the negative power by moving it downstairs)}\\ y^{\prime }=\frac{1}{\ln 7}\cdot \frac{1}{2}\cdot \left(4\right)\left[\frac{1}{{\left(4x-3\right)}^{1/2}}\cdot \frac{1}{{\left(4x-3\right)}^{1/2}}\right]\text{(rearranged factors using associative and commutative laws)}\\ y^{\prime }=\frac{4}{2\ln 7}\left[\frac{1}{4x-3}\right]=\underset{¯}{\underset{¯}{\frac{2}{\left(\ln 7\right)\left(4x-3\right)}}}\text{(same result, but more work; LEARN YOUR LOGS!)}\end{array}$$

4.5 #39.

$$\begin{array}{l}z=\left({10}^y\right)\left({\log }_{10}y\right)\text{This is a product rule situation with both exponential and logarithmic functions (no chain rule!)}\text{.}\\ u={10}^y;u^{\prime }=\left({10}^y\right)\left(\ln 10\right)\text{(the rule for exponential bases other than}e\text{is in the text, p}\text{. 274 at bottom; use}a=10\text{)}\\ v=\left({\log }_{10}y\right);v^{\prime }=\frac{1}{y}\cdot \frac{1}{\ln 10}\text{(the extra fraction on the end is part of the derivative of the log, not a chain rule)}\\ z^{\prime }=u^{\prime }v+v^{\prime }u=\left({10}^y\right)\left(\ln 10\right)\left({\log }_{10}y\right)+\left(\frac{1}{y}\cdot \frac{1}{\ln 10}\right)\left({10}^y\right)=\underset{¯}{\underset{¯}{\left({10}^y\right)\left(\ln 10\right)\left({\log }_{10}y\right)+\frac{{10}^y}{\left(\ln 10\right)y}}}\\ \text{(This is the same as the text answer; the order of terms is a little different}\text{. I don't know why they didn't factor out the}{10}^y\text{.)}\end{array}$$

4.5 #41.

$$\begin{array}{l}f(x)=\ln \left(x{e}^{\sqrt{x}}+2\right)\text{This is a logarithm containing a product, with two nested layers of chain rule inside}\text{.}\\ \text{It is also a good place to try the}\exp (x)\text{notation that I mentioned in class as an alternative to}{e}^x\text{.}\\ \text{Prepare the derivative of}x{e}^{\sqrt{x}}=x\cdot \exp \left(\sqrt{x}\right)=x\cdot \exp \left(x^{1/2}\right)\text{inside the parentheses for later use, using the product rule}\text{.}\\ u=x;u^{\prime }=1\\ v=\exp \left(x^{1/2}\right);v^{\prime }=\exp \left(x^{1/2}\right)\cdot \frac{d}{dx}\left(x^{1/2}\right)=\exp \left(x^{1/2}\right)\cdot \frac{1}{2}{x}^{-1/2}=\exp \left(x^{1/2}\right)\cdot \frac{1}{2x^{1/2}}=\frac{\exp \left(x^{1/2}\right)}{2x^{1/2}}\\ \frac{d}{dx}\left(x{e}^{\sqrt{x}}\right)=u^{\prime }v+v^{\prime }u=\left(1\right)\left[\exp \left(x^{1/2}\right)\right]+\left[\frac{\exp \left(x^{1/2}\right)}{2x^{1/2}}\right]\left(x\right)=\exp \left(x^{1/2}\right)+\frac{x^1\exp \left(x^{1/2}\right)}{2x^{1/2}}\\ \frac{d}{dx}\left(x{e}^{\sqrt{x}}\right)=\left[\exp \left(x^{1/2}\right)\right]\left[1+\frac{x^1}{2x^{1/2}}\right]\\ \frac{d}{dx}\left(x{e}^{\sqrt{x}}\right)=\left[\exp \left(x^{1/2}\right)\right]\left(\frac{1}{1}+\frac{x^{1/2}}{2}\right)\text{(prepare the 1 for LCDs and reduce the last fraction)}\\ \frac{d}{dx}\left(x{e}^{\sqrt{x}}\right)=\left[\exp \left(x^{1/2}\right)\right]\left(\frac{2}{2}+\frac{x^{1/2}}{2}\right)=\left[\exp \left(x^{1/2}\right)\right]\left(\frac{2+x^{1/2}}{2}\right)=\left[\exp \left(x^{1/2}\right)\right]\cdot \frac{1}{2}\left(2+x^{1/2}\right)\\ \frac{d}{dx}\left(x{e}^{\sqrt{x}}\right)=\frac{1}{2}{e}^{\sqrt{x}}\left(2+\sqrt{x}\right)\text{(and now, back to our regularly scheduled derivative)}\\ f^{\prime }(x)=\frac{1}{\left(x{e}^{\sqrt{x}}+2\right)}\cdot \left[\frac{d}{dx}\left(x{e}^{\sqrt{x}}+2\right)\right]\text{(log rule, and chain rule on the stuff inside the log)}\\ f^{\prime }(x)=\frac{1}{\left(x{e}^{\sqrt{x}}+2\right)}\cdot \left[\frac{d}{dx}\left(x{e}^{\sqrt{x}}\right)+\frac{d}{dx}\left(2\right)\right]\text{(sum rule inside the brackets; last term is of course zero)}\\ f^{\prime }(x)=\frac{1}{\left(x{e}^{\sqrt{x}}+2\right)}\cdot \left[\frac{1}{2}{e}^{\sqrt{x}}\left(2+\sqrt{x}\right)\right]\text{(plug in the derivative of}x{e}^{\sqrt{x}}\text{found above)}\\ f^{\prime }(x)=\underset{¯}{\underset{¯}{\frac{e^{\sqrt{x}}\left(2+\sqrt{x}\right)}{2\left(x{e}^{\sqrt{x}}+2\right)}}}\end{array}$$

4.5 #43.

$$\begin{array}{l}f(t)=\frac{2t^{3/2}}{\ln \left(2t^{3/2}+1\right)}\text{This is a quotient situation; the logarithm also contains a chain rule; serious algebra awaits us}\text{.}\\ \text{Logarithm "gotcha" for later: Note that}\ln \left(x^2\right)\ne {\left(\ln x\right)}^2\text{; the left side can be changed to}2\ln x,\text{but not the right side!}\\ u=2t^{3/2};u^{\prime }=\left(2\right)\left(\frac{3}{2}{t}^{1/2}\right)=3t^{1/2}\\ v=\ln \left(2t^{3/2}+1\right);v^{\prime }=\frac{1}{\left(2t^{3/2}+1\right)}\cdot \frac{d}{dt}\left(2t^{3/2}+1\right)=\frac{1}{\left(2t^{3/2}+1\right)}\cdot \left(3t^{1/2}\right)={\left(2t^{3/2}+1\right)}^{-1}\left(3t^{1/2}\right)\\ f^{\prime }(t)=\frac{u^{\prime }v-v^{\prime }u}{v^2}=\frac{\left(3t^{1/2}\right)\left[\ln \left(2t^{3/2}+1\right)\right]-\left[{\left(2t^{3/2}+1\right)}^{-1}\left(3t^{1/2}\right)\right]\left(2t^{3/2}\right)}{{\left[\ln \left(2t^{3/2}+1\right)\right]}^2}\\ f^{\prime }(t)=\frac{\left(3t^{1/2}\right)\left[\ln \left(2t^{3/2}+1\right)\right]-{\left(2t^{3/2}+1\right)}^{-1}\left[\left(3t^{1/2}\right)\left(2t^{3/2}\right)\right]}{{\left[\ln \left(2t^{3/2}+1\right)\right]}^2}\text{(apply associative law to second term upstairs)}\\ f^{\prime }(t)=\frac{{\left(2t^{3/2}+1\right)}^1}{{\left(2t^{3/2}+1\right)}^1}\cdot \frac{\left\{\left(3t^{1/2}\right)\left[\ln \left(2t^{3/2}+1\right)\right]-{\left(2t^{3/2}+1\right)}^{-1}\left(6t^2\right)\right\}}{{\left[\ln \left(2t^{3/2}+1\right)\right]}^2}\text{(multiply top and bottom by the fraction in front)}\\ f^{\prime }(t)=\frac{\left(2t^{3/2}+1\right)\left(3t^{1/2}\right)\left[\ln \left(2t^{3/2}+1\right)\right]-{\left(2t^{3/2}+1\right)}^1{\left(2t^{3/2}+1\right)}^{-1}\left(6t^2\right)}{\left(2t^{3/2}+1\right){\left[\ln \left(2t^{3/2}+1\right)\right]}^2}\text{(distribute the new factors)}\\ f^{\prime }(t)=\underset{¯}{\underset{¯}{\frac{\left(6t^2+3t^{1/2}\right)\left[\ln \left(2t^{3/2}+1\right)\right]-6t^2}{\left(2t^{3/2}+1\right){\left[\ln \left(2t^{3/2}+1\right)\right]}^2}}}\text{(upstairs: distribute the first term, and cancel opposite powers in the second term)}\end{array}$$