Mary Monroe-Ellis
Susan Mosteller
The Gross Domestic Product (GDP) of the U.S. (in billions of dollars) t years after the year 2000 can be modeled by $G(t)=9743.77{{e}^{0.0514t}}$
$G(0)=9743.77{{e}^{0.0514(0)}}$
$=9743.77{{e}^{0}}$
$=9743.77(1)$
$=9743.77$
In the year 2000, the US GDP was $\$9743.77$ billion.
$G(7)=9743.77{{e}^{0.0514(7)}}=\$13,963.24$
According to the model, GDP in 2007 should have been $\$13,963.24$
$G(10)=9743.77{{e}^{0.0514(10)}}=\$16,291.25$
According to the model, GDP in 2010 should have been $\$16,291.25.$
Solve the equations. Round the answers to two decimal places.
$4={{e}^{0.08t}}$
$\ln 4=\ln {{e}^{0.08t}}$
$\ln 4=(1)0.08t$
$\ln 4=0.08t$
$\frac{\ln 4}{0.08}=t$
$17.33=t$
$A=2500{{e}^{0.5(20)}}$
$A={{2500}{{e}^{10}}}$
$A=55,066,164.49$
$20000=P{{e}^{0.0125(6)}}$
$20,000=P{{e}^{0.075}}$
$\frac{20,000}{{{e}^{0.075}}}=P$
$18,554.87=P$
$15000=12500{{e}^{0.005t}}$
$\frac{15000}{12500}=\frac{12500{{e}^{0.005t}}}{12500}$
$1.2={{e}^{0.005t}}$
$\ln 1.2=\ln {{e}^{0.005t}}$
$\ln 1.2=0.005t$
$\frac{\ln 1.2}{0.005}=t$
$36.46=t$
$5625=4850{{e}^{5.25x}}$
$\frac{5625}{4850}=\frac{4850{{e}^{5.25x}}}{4850}$
$\frac{225}{194}={{e}^{5.25x}}$
$\ln \frac{225}{194}=\ln {{e}^{5.25x}}$
$\ln \frac{225}{194}=5.25x$
$\left( \ln \frac{225}{194} \right)/5.25=x$
$0.03=x$
$y=\ln {{e}^{7}}$
$y=7\ln e$
$y=7(1)$
$y=7$
${{\log }_{3}}(7-2x)=2$
${{3}^{2}}=7-2x$
$9=7-2x$
$2=-2x$
$-1=x$
$\ln \ ({{x}^{2}}-99)=0$
${{e}^{0}}={{x}^{2}}-99$
$1={{x}^{2}}-99$
$0={{x}^{2}}-100$
$0=(x+10)(x-10)$
$x=\pm 10$
$\ln \ ({{x}^{2}}-20)=\ln \ (x)$
${{x}^{2}}-20=x$
${{x}^{2}}-x-20=0$
$(x-5)(x+4)=0$
$x=5\quad \quad x=-4$
$x=-4$ is an extraneous root
The only solution is $x=5.$
Answer the following questions. Show all of your work. Round to two decimal places.
If you invested $1,000 in an account paying an annual percentage rate (quoted rate) of 2%, compounded continuously, how much would you have in your account at the end of
$A = 1000{e^{.02(1)}} = 1020.20$
At the end of one year, there will be \$1020.20 in the account.
$A = 1000{e^{.02(10)}} = 1221.40$
At the end of ten years, there will be \$1221.40 in the account.
$A = 1000{e^{.02(20)}} = 1491.82$
At the end of twenty years, there will be \$1491.82 in the account.
$A = 1000{e^{.02(50)}} = 2718.28$
At the end of fifty years, there will be \$2718.28 in the account.
A $1,000 investment is made in a trust fund at an annual percentage rate of 12%, compounded continuously. How long will it take the investment to
$2000 = 1000{e^{0.12t}}$
$2 = {e^{0.12t}}$
$\ln 2 = \ln {e^{0.12t}}$
$\ln 2 = 0.12t$
$\frac{{\ln 2}}{{0.12}} = t$
$t = 5.78$
The investment will double in 5.78 years.
$3000 = 1000{e^{0.12t}}\quad$
$3 = {e^{0.12t}}$
$\ln 3 = \ln {e^{0.12t}}$
$\ln 3 = 0.12t$
$\frac{{\ln 3}}{{0.12}} = t$
$t = 9.16$
The investment will triple in 9.16 years.
If $500 is invested in an account which offers 0.75%, compounded continuously find:
$A = 500{e^{.0075t}}$
$A(5) = 500{e^{.0075(5)}}=519.11$
After 5 years, \$519.11 will be in the account.
$A(10) = 500{e^{.0075(10)}}=538.94$
After 10 years, \$538.94 will be in the account.
$A(30) = 500{e^{.0075(30)}}=626.16$
After 30 years, \$626.16 will be in the account.
$A(35) = 500{e^{.0075(35)}}=650.09$
After 35 years, \$650.09 will be in the account.
$1000 = 500{e^{.0075t}}$
$2 = {e^{.0075t}}$
$\ln 2 = \ln {e^{.0075t}}$
$\ln 2 = .0075t$
$\frac{{\ln 2}}{{.0075}} = t$
It will take 92.42 years for the initial investment to double.
4th year $A(4) = 500{e^{.0075(4)}} = \$515.23$
5th year $A(5) = 500{e^{.0075(5)}} = \$519.11$
$\frac{A(5)-A(4)}{5-4}=\frac{519.11-515.23}{1}=3.88$
The balance in the account is increasing by an average of $3.88 per year.
If $5000 is invested in an account which offers 2.125%, compounded continuously, find:
$A(t) = 5000{e^{.02125t}}$
$A(5) = 5000{e^{.02125(5)}} = 5560.50$
After 5 years, \$5560.50 will be in the account.
$A(10) = 5000{e^{.02125(10)}} = 6183.83$
After 10 years, \$6183.83 will be in the account.
$A(30) = 5000{e^{.02125(30)}} = 9458.73$
After 30 years, \$9458.73 will be in the account.
$A(35) = 5000{e^{.02125(35)}} =10519.05$
After 35 years, \$10519.05 will be in the account.
$10,000 = 5000{e^{.02125t}}$
$2 = {e^{.02125t}}$
$\ln 2 = \ln {e^{.02125t}}$
$\ln 2 = .02125t$
$\frac{{\ln 2}}{{.02125}} = t$
It will take 32.62 years for the initial investment to double
$A(4) = 5000{e^{.02125(4)}}$
$A(4) = \$ 5443.59$
$A(5) = 5000{e^{.02125(5)}}$
$A(5) = \$ 5560.50$
$A(5)-A(4) = 5560.50-5443.59=116.91$
The balance in the account is increasing by an average of $116.91 per year.
How much money needs to be invested now to obtain \$5000 in 10 years if the interest rate in a CD is 2.25%, compounded continuously?
$A = P{e^{rt}}$
$5000 = P{e^{0.0225(10)}}$
$5000 = P{e^{0.225}}$
$\frac{{5000}}{{{e^{0.225}}}} = \frac{{P{e^{0.225}}}}{{{e^{0.225}}}}$
$\$ 3992.58 = P$
\$3992.58 needs to be invested now, in order to have $5000 in 10 years.
A mathematical model for depreciation of a car is given by $A = P{(1-r)^t}$, where A is defined as the value of the car after t years, P is defined as the original value of the car, and r is the rate of depreciation per year. The cost of a new car is $32,000. It depreciates at a rate of 15% per year. This means that it loses 15% of its value each year.
$A = 32,000{\left( {1-0.15} \right)^t}$
$A = 32,000{\left( {0.85} \right)^t}$
$A = 32,000{\left( {.85} \right)^4}$
$A = 16,704.20$
The car is worth \$ 16,704.20 when it is 4 years old.
A mathematical model for depreciation of an ATV (all-terrain vehicle) is given by $A = P{(1-r)^t}$, where A is defined as the value of the vehicle after t years, P is defined as the original value of the vehicle, and r is the rate of depreciation per year. The cost of a new ATV (all-terrain vehicle) is \$7200. It depreciates at 18% per year.
$A = 7200{\left( {1-0.18} \right)^t}$
$A = 7200{\left( {.82} \right)^t}$
$A = 7200{\left( {.82} \right)^{10}}$
$A = 989.63$
The value of the ATV when it is 10 years old will be \$989.63.
Michigan’s population is declining at a rate of 0.5% per year. In 2004, the state had a population of 10,112,620.
$y = 10,112,620{\left( {.995} \right)^t}$
$y = 10,112,620{\left( {.995} \right)^8}$
In 2012 the population of Michigan will be approximately 9,715,124 people.
$9,900,000 = 10,112,620{\left( {.995} \right)^t}$
$\frac{{9,900,000}}{{10,112,620}} = {.995^t}$
$\ln (\frac{{9,9000,000}}{{10,112,620}}) = \ln {.995^t}$
$\ln (\frac{{9,900,000}}{{10,112,620}}) = t\;\ln .995$
$\frac{{\ln (\frac{{9,900,000}}{{10,112,60}})}}{{\ln .995}} = t$
$t = 4.24$ years
The population of Michigan will be 9,900,000 people in March of 2008.
$y = 10,112,620{\left( {.995} \right)^{ - 4}}$
$y = 10,317,426.06$
According to this model, the population of Michigan in 2000 was 10,317,426 people.
https://sccmath.files.wordpress.com/2012/01/scc_open_source_intermediate_algebra.pdf
Find the derivative of each of the following.
$f(x) = {e^x} + \ln x - 2{x^5} + 12$
$f'(x) = {e^x} + \frac{1}{x} - 10{x^4}$
$y = 7\ln x + 14x - \frac{1}{2}$
${y'} = \frac{7}{x} + 14$
$g(x) = - 5{e^x} - 12\ln x + 3{x^3}$
$g'(x) = - 5{e^x} - \frac{{12}}{x} + 9{x^2}$
$f(x) = 8\sqrt x + 7{e^x}$
$f(x) = 8{x^{1/2}} + 7{e^x}$
$f'(x) = 4{x^{ - 1/2}} + 7{e^x}$
$f'(x) = \frac{4}{{\sqrt x }} + 7{e^x}$
$y = \ln {x^7} - 3\ln x$
$y = 7\ln x - 3\ln x$
$y' = \frac{7}{x} - \frac{3}{x}$
$y' = \frac{4}{x}$
$f(x) = 4\ln \frac{1}{x} + 8$
$f(x) = 4(\ln 1 - \ln x) + 8$
$f(x) = 4\ln 1 - 4\ln x + 8$
$f'(x) = 0 - \frac{4}{x} + 0$
$f'(x) = \frac{{ - 4}}{x}$
$y = {e^x} - 7\ln 5x + 14$
$y = {e^x} - 7(\ln 5 + \ln x) + 14$
$y = {e^x} - 7\ln 5 - 7\ln x + 14$
$y' = {e^x} - 0 - \frac{7}{x} + 0$
$y' = {e^x} - \frac{7}{x}$
Find the equation of the line tangent to the graph of f at the indicated value of x.
$f(x) = 2{e^x} - 1\quad \quad at\quad x = 0$
$f(0) = 2{e^0} - 1$
$f(0) = 2(1) - 1 = 1$
$(0,1)$
$f'(x) = 2{e^x}$
${m_{\tan }} = f'(0) = 2{e^0} = 2(1) = 2$
$y - 1 = 2(x - 0)$
$y - 1 = 2x$
$y = 2x + 1$
Find the equation of the line tangent to the graph of f at the indicated value of x.
$f(x) = 8\ln (x)\quad \quad at\quad x = e$
$f(e) = 8\ln e$
$f(e) = 8(1) = 8$
$(e,8)$
$f'(x) = \frac{8}{x}$
${m_{\tan }}=f'(e) = \frac{8}{e}$
$y - 8 = \frac{8}{e}(x - e)$
$y - 8 = \frac{8}{e}x - 8$
$y = \frac{{8x}}{e}$
An editor of college textbooks has determined that the equation below models the sales of a calculus textbook B (in thousands) based on the number of complimentary books sent to professors x (also in thousands). $$B(x) = 3.24 + 1.6\ln \;(x)$$ Find and interpret the instantaneous rate of change when x=6.
$B'(x) = 0 + \frac{{1.6}}{x} = \frac{{1.6}}{x}$
$B'(6) = \frac{{1.6}}{6} = .2\overline {6} \approx .27$
When 6000 complimentary books are sent to professors, the sales of textbooks increase by 270 (.27 thousand) books.
The percentage of mothers who returned to the work force within one year after they had a child for the years 1976 through 1998 can be modeled by $$P(t) = 36.025 + 6.27\ln (t)$$ percent t years after 1977. (Source: Based on data from the Associated Press)
$t = 21\quad $when the year is 1998
$P(21) = 36.025 + 6.27\;\ln (21) = 55.11\%$
55.11% of mothers returned to the work force within one year in 1998.
$P'(21) = \frac{{6.27}}{t} = \frac{{6.27}}{{21}} = 0.30\%$
The percentage of mothers returning to work within one year was increasing at a rate of $0.30\%$.
$\frac{{P(13) - P(3)}}{{13 - 3}} = \frac{{52.11 - 42.91}}{{13 - 3}} = \frac{{9.2}}{{10}} = .92\% $
The percent of mothers returning to the work force increased on average .92% per year.
c. What happens to the rate at which the percentage is growing as more years go by?As more years go by, the change in percentage growns at a slower rate.
http://www2.fiu.edu/~rosentha/MAC2233/2233CE.htm (page 7 Section 3.3)
Find the derivative of each of the following.
$p(t) = 2{t^2}({t^3} + 4t)$
$f = 2{t^2}$
$f' = 4t$
$g = {t^3} + 4t$
$g' = 3{t^2} + 4$
$p'(t) = 4t({t^3} + 4t) + 2{t^2}(3{t^2} + 4)$
$p'(t) = 4{t^4} + 16{t^2} + 6{t^4} + 8{t^2}$
$p'(t) = 10t{}^4 + 24{t^2}$
$k(x) = (2x - 5)({x^2} + 1)$
$f = 2{t^2}$
$f' = 4t$
$g = {t^3} + 4t$
$g' = 3{t^2} + 4$
$k'(x) = 2({x^2} + 1) + 2x(2x - 5)$
$k'(x) = 2{x^2} + 2 + 4{x^2} - 10x$
$k'(x) = 6{x^2} - 10x + 2$
$h(x) = \frac{{6x + 5}}{{3x - 8}}$
$f = 6x + 5$
$f' = 6$
$g = 3x - 8$
$g' = 3$
$h'(x) = \frac{{6(3x - 8) - 3(6x + 5)}}{{{{\left( {3x - 8} \right)}^2}}}$
$h'(x) = \frac{{18x - 48 - 18x - 15}}{{{{\left( {3x - 8} \right)}^2}}}$
$h'(x) = \frac{{ - 63}}{{{{\left( {3x - 8} \right)}^2}}}$
$p(a) = ({a^2} - 2a + 7)(2{a^2} - a + 1)$
$f = {a^2} - 2a + 7$
$f' = 2a - 2$
$g = 2{a^2} - a + 1$
$g' = 4a - 1$
$p'(a) = (2a - 2)(2{a^2} - a + 1) + (4a - 1)({a^2} - 2a + 7)$
$p'(a) = 4{a^3} - 2{a^2} + 2a - 4{a^2} + 2a - 2 + 4{a^3} - 8{a^2} + 28a - {a^2} + 2a - 7$
$p'(a) = 8{a^3} - 15{a^2} + 34a - 9$
$y = \frac{{{x^3} - 5x}}{{4{x^2}}}$
$f = {x^3} - 5x$
$f' = 3{x^2} - 5$
$g = 4{x^2}$
$g' = 8x$
$y' = \frac{{(3{x^2} - 5)(4{x^2}) - 8x({x^3} - 5x)}}{{{{\left( {4{x^2}} \right)}^2}}}$
$y' = \frac{{12{x^4} - 20{x^2} - 8{x^4} + 40{x^2}}}{{{{\left( {4{x^2}} \right)}^2}}}$
$y' = \frac{{4{x^4} + 20{x^2}}}{{16{x^4}}}$
$y' = \frac{{4{x^2}({x^2} + 5)}}{{16{x^4}}}$
$y' = \frac{{{x^2} + 5}}{{4{x^2}}}$
$p(x) = 2{e^x}({x^2} - 3x + 5)$
$f = 2{e^x}$
$f' = 2{e^x}$
$g = {x^2} - 3x + 5$
$g' = 2x - 3$
$p'(x) = 2{e^x}({x^2} - 3x + 5) + 2{e^x}(2x - 3)$
$p'(x) = 2{e^x}({x^2} - 3x + 5 + 2x - 3)$
$p'(x) = 2{e^x}({x^2} - x + 2)$
$q(t) = \frac{{\ln x}}{{{x^4} - 5}}$
$f = \ln x$
$f' = \frac{1}{x}$
$g = {x^4} - 5$
$g' = 4{x^3}$
$q'(t) = \frac{{\frac{1}{x}({x^4} - 5) - 4{x^3}\ln x}}{{{{\left( {{x^4} - 5} \right)}^2}}}$
$q'(t) = \frac{{{x^3} - \frac{5}{x} - 4{x^3}\ln x}}{{{{\left( {{x^4} - 5} \right)}^2}}}$
Multiply $q'(t)$ by $\frac{x}{x}$
$q'(t) = \frac{{{x^4} - 5 - 4{x^4}\ln x}}{{x{{\left( {{x^4} - 5} \right)}^2}}}$
The concentration of a particular drug in the bloodstream can be modeled by $C(t) = \frac{{20t}}{{{t^2} + t + 5}}$, where $C(t)$ is measured in milligrams per cubic millimeter and $t$ is time in hours. What change in concentration can be expected between the 5th and 6th hour after the drug is administered?
$f = 20t$
$f' = 20$
$g = {t^2} + t + 5$
$g' = 2t +1$
$C'(t) = \frac{{20({t^2} + t + 5) - 20t(2t + 1)}}{{{{\left( {{t^2} + t + 5} \right)}^2}}}$
$C'(t) = \frac{{20{t^2} + 20t + 100 - 40{t^2} - 20t}}{{{{\left( {{t^2} + t + 5} \right)}^2}}}$
$C'(t) = \frac{{ - 20{t^2} + 100}}{{{{\left( {{t^2} + t + 5} \right)}^2}}}$
$C'(5) = \frac{{ - 20{{(5)}^2} + 100}}{{{{\left( {{5^2} + 5 + 5} \right)}^2}}}$
$C'(5) = - 0.33$
The concentration of the drug is decreasing by .33 mg/ml$^3$ from the 5th to 6th hours.
Determine the slope of the tangent line to the curve $R(x) = \frac{{{x^2} - 2x - 8}}{{{x^2} - 9}}$ at the point where $x = 0$.
$f = {x^2} - 2x - 8$
$f' = 2x - 2$
$g = {x^2} - 9$
$g' = 2x$
$R'(x) = \frac{{(2x - 2)({x^2} - 9) - 2x({x^2} - 2x - 8)}}{{{{\left( {{x^2} - 9} \right)}^2}}}$
$R'(x) = \frac{{2{x^3} - 18x - 2{x^2} + 18 - 2{x^3} + 4{x^2} + 16x}}{{{{\left( {{x^2} - 9} \right)}^2}}}$
$R'(x) = \frac{{2{x^2} - 2x + 18}}{{{{\left( {{x^2} - 9} \right)}^2}}}$
${m_{\tan }}=R'(0) = \frac{{2{{(0)}^2} - 2(0) + 18}}{{{{\left( {{0^2} - 9} \right)}^2}}} = \frac{{18}}{{81}} = \frac{2}{9}$
${m_{\tan }} =\frac{2}{9}$
When a camera flashes, the intensity $I$ of light seen by the eye is given by the function $I(t) = \frac{{100t}}{{{e^t}}}$ where $I$ is measured in candles and $t$ is measured in milliseconds. Compute $I'(0.5)$, $I'(2)$, and $ I'(5)$; include appropriate units on each value; and discuss the meaning of each.
$f = 100t$
$f' = 100$
$g = {e^t}$
$g' = {e^t}$
$I'(t) = \frac{{100{e^t} - 100t{e^t}}}{{{{\left( {{e^t}} \right)}^2}}}$
$I'(t) = \frac{{{e^t}(100 - 100t)}}{{{e^{2t}}}}$
$I'(t) = \frac{{100 - 100t}}{{{e^t}}}$
$I'(0.5) =30.33$
The intensity of the light seen by the eye 0.5 milliseconds after the flash is increasing by 30.33 candles per millisecond.
$I'(2) = -13.53$
The intensity of the light seen by the eye 2 milliseconds after the flash is decreasing by 13.53 candles per millisecond.
$I'(5) = -2.70$
The intensity of the light seen by the eye 5 milliseconds after the flash is decreasing by 2.70 candles per millisecond.
Active Calculus by Matthew Boelkins is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. Based on a work at http://scholarworks.gvsu.edu/books/10/./
Find the derivative of each of the following.
$p(t) = \sqrt t \;({t^2} + 5t - 2)$
$f = {t^{1/2}}$
$f' = \frac{1}{2}{t^{ - 1/2}}$
$g = {t^2} + 5t - 2$
$g' = 2t + 5$
$p'(t)=\frac{1}{2}{{t}^{-1/2}}({{t}^{2}}+5t-2)+{{t}^{1/2}}(2t+5)$
$p'(t)=\frac{1}{2}{{t}^{3/2}}+\frac{5}{2}{{t}^{{}^{1}/{}_{2}}}-{{t}^{-1/2}}+2{{t}^{3/2}}+5{{t}^{1/2}}$
$p'(t)=\frac{5}{2}{{t}^{3/2}}+\frac{15}{2}{{t}^{1/2}}-{{t}^{-1/2}}$
$p'(t)=\frac{5}{2}{{t}^{3/2}}+\frac{15}{2}{{t}^{1/2}}-{\frac{1}{{t}^{1/2}}}$
$r(x) = {x^{ - 4}}({x^2} - 7x + 1)$
$f = {x^{ - 4}}$
$f' = - 4{x^{ - 5}}$
$g = {x^2} - 7x + 1$
$g' = 2x - 7$
$r'(x) = - 4{x^{ - 5}}({x^2} - 7x + 1) + {x^{ - 4}}(2x - 7)$
$r'(x) = - 4{x^{ - 3}} + 28{x^{ - 4}} - 4{x^{ - 5}} + 2{x^{ - 3}} - 7{x^{ - 4}}$
$r'(x) = - 2{x^{ - 3}} + 21{x^{ - 4}} - 4{x^{ - 5}}$
$m(x) = \frac{{\sqrt x }}{{3x - 8}}$
$f = {x^{1/2}}$
$f' = \frac{1}{2}{x^{ - 1/2}}$
$g = 3x - 8$
$g' = 3$
$m'(x)=\frac{\frac{1}{2}{{x}^{-1/2}}(3x-8)-3{{x}^{1/2}}}{{{\left( 3x-8 \right)}^{2}}}$
$m'(x)=\frac{\frac{3}{2}{{x}^{1/2}}-4{{x}^{-1/2}}-3{{x}^{1/2}}}{{{\left( 3x-8 \right)}^{2}}}$
$m'(x)=\frac{\frac{-3}{2}{{x}^{1/2}}-4{{x}^{-1/2}}}{{{\left( 3x-8 \right)}^{2}}}$
$m'(x)=\frac{\frac{-3{{x}^{1/2}}}{2}-\frac{4}{{{x}^{1/2}}}}{{{\left( 3x-8 \right)}^{2}}}$
Multiply $m'(x)$ by $\frac{x^{1/2}}{x^{1/2}}$
$m'(x)=\frac{\frac{-3x}{2}-4}{{{x}^{{}^{1}/{}_{2}}}{{\left( 3x-8 \right)}^{2}}}$
Multiply $m'(x)$ by $\frac{2}{2}$
$m'(x)=\frac{-3x-8}{2\sqrt{x}{{\left( 3x-8 \right)}^{2}}}$
$p(a) = ({e^a} + 1)(\sqrt[3]{{{a^2}}} + 2)$
$f = {e^a} + 1$
$f' = {e^a}$
$g = {a^{2/3}} + 2$
$g' = \frac{2}{3}{a^{ - 1/3}}$
$p'(a) = {e^a}({a^{2/3}} + 2) + \frac{2}{3}{a^{ - 1/3}}({e^a} + 1)$
$p'(a)={{e}^{a}}{{a}^{2/3}}+2{{e}^{a}}+\frac{2}{3}{{e}^{a}}{{a}^{-1/3}}+\frac{2}{3}{{a}^{-1/3}}$
$y=\frac{x^{\frac15}-x}{x^{\frac25}}$
$f={{x}^{1/5}}-x$
$f'=\frac{1}{5}{{x}^{-4/5}}-1$
$g={{x}^{2/5}}$
$g'=\frac{2}{5}{{x}^{-3/5}}$
$y'=\frac{\left( \frac{1}{5}{{x}^{-4/5}}-1 \right){{x}^{2/5}}-\frac{2}{5}{{x}^{-3/5}}\left( {{x}^{1/5}}-x \right)}{{{\left( {{x}^{2/5}} \right)}^{2}}}$
$y'=\frac{\frac{1}{5}{{x}^{-2/5}}-{{x}^{2/5}}-\frac{2}{5}{{x}^{-2/5}}+\frac{2}{5}{{x}^{2/5}}}{{{x}^{4/5}}}$
Multiply $y'$ by $\frac{5}{5}$
$y'=\frac{{{x}^{-2/5}}-5{{x}^{2/5}}-2{{x}^{-2/5}}+2{{x}^{2/5}}}{5{{x}^{4/5}}}$
$y'=\frac{-1{{x}^{-2/5}}-3{{x}^{2/5}}}{5{{x}^{4/5}}}$
$y'=\frac{\frac{-1}{{{x}^{{}^{2}/{}_{5}}}}-3{{x}^{2/5}}}{5{{x}^{4/5}}}$
Multiply $y'$ by $\frac{x^{{}^{2}/{}_{5}}}{x^{{}^{2}/{}_{5}}}$
$y'=\frac{-1-3{{x}^{4/5}}}{5{{x}^{6/5}}}$
$h(x) = (2 + \ln x)( - 4{x^3} + {e^x})$
$f = 2 + \ln x$
$f' = \frac{1}{x}$
$g = - 4{x^3} + {e^x}$
$g' = - 12{x^2} + {e^x}$
$h'(x) = \frac{1}{x}( - 4{x^3} + {e^x}) + (2 + \ln x)( - 12{x^2} + {e^x})$
$h'(x) = - 4{x^2} + \frac{{{e^x}}}{x} - 24{x^2} + 2{e^x} - 12{x^2}\ln x + {e^x}\ln x$
$h'(x) = - 28{x^2} + \frac{{e^x}}{x} + 2{e^x} - 12{x^2}\ln x + {e^x}\ln x$
$q(t) = \frac{{\sqrt[5]{{{t^3}}}}}{{7\ln t}}$
$f = {t^{3/5}}$
$f' = \frac{3}{5}{t^{ - 2/5}}$
$g = 7\ln t$
$g' = \frac{7}{t}$
$q'(t)=\frac{\frac{3}{5}{{t}^{-2/5}}(7\ln t)-\frac{7}{t}({{t}^{3/5}})}{{{\left( 7\ln t \right)}^{2}}}$
$q'(t)=\frac{\frac{21}{5}{{t}^{-2/5}}\ln t-7{{t}^{-2/5}}}{49{{\left( \ln t \right)}^{2}}}$
Multiply $q'(t)$ by $\frac{5{{t}^{2/5}}}{5{{t}^{2/5}}}$
$q'(t)=\frac{21\ln t-35}{{245{{t}^{2/5}}}{{\left( \ln t \right)}^{2}}}$
$q'(t)=\frac{7(3\ln t-5)}{245{{t}^{2/5}}{{\left( \ln t \right)}^{2}}}$
$q'(t)=\frac{3\ln t-5}{35{{t}^{2/5}}{{\left( \ln t \right)}^{2}}}$
Find the equation of the tangent line to $f(x) = \frac{{{x^2} + 3}}{{6 - x}}$ at $x = 3$.
$f(3) = \frac{{{3^2} + 3}}{{6 - 3}} = \frac{{9 + 3}}{3} = \frac{{12}}{3} = 4$
$(3,4)$
$h = {x^2} + 3$
$h' = 2x$
$g = 6 - x$
$g' = - 1$
$f'(x) = \frac{{2x(6 - x) + 1({x^2} + 3)}}{{{{\left( {6 - x} \right)}^2}}}$
$f'(x) = \frac{{12x - 2{x^2} + {x^2} + 3}}{{{{\left( {6 - x} \right)}^2}}}$
$f'(x) = \frac{{ - 1{x^2} + 12x + 3}}{{{{\left( {6 - x} \right)}^2}}}$
${m_{\tan }}= f'(3) = \frac{{ - 1({3^2}) + 12(3) + 3}}{{{{\left( {6 - 3} \right)}^2}}}$
${m_{\tan }}= \frac{{ - 9 + 36 + 3}}{{{3^2}}} = \frac{{30}}{9} = \frac{{10}}{3}$
$y - 4 = \frac{{10}}{3}(x - 3)$
$y - 4 = \frac{{10}}{3}x - 10$
$y = \frac{{10}}{3}x - 6$
A music publisher expects that over the first 24 months after the release of an album the monthly profit (in thousands of dollars) can be modeled by $P(t) = \frac{{800 + 400t - 20{t^2}}}{{{t^2} + 20}}$
$f = 800 + 400t - 20{t^2}$
$f' = 400 - 40t$
$g = {t^2} + 20$
$g' = 2t$
$P'(t) = \frac{{(400 - 40t)({t^2} + 20) - 2t(800 + 400t - 20{t^2})}}{{{{\left( {{t^2} + 20} \right)}^2}}}$
$P'(t) = \frac{{400{t^2} + 8000 - 40{t^3} - 800t - 1600t - 800{t^2} + 40{t^3}}}{{{{\left( {{t^2} + 20} \right)}^2}}}$
$P'(t) = \frac{{ - 400{t^2} - 2400t + 8000}}{{{{\left( {{t^2} + 20} \right)}^2}}}$
$P'(1) = \frac{{ - 400{{(1)}^2} - 2400(1) + 8000}}{{{{\left( {{1^2} + 20} \right)}^2}}}$
$P'(1) = \frac{{ - 400 - 2400 + 8000}}{{{{\left( {21} \right)}^2}}} = \frac{{5200}}{{441}} = 11.79$
The profit is increasing at a rate of \$11,790 per month after the 1st month of release.
$P'(6) = \frac{{ - 400{{(6)}^2} - 2400(6) + 8000}}{{{{\left( {{6^2} + 20} \right)}^2}}}$
$P'(6) = \frac{{ - 14400 - 14400 + 8000}}{{{{56}^2}}} = \frac{{ - 20800}}{{3136}} = - 6.63$
The profit is decreasing at a rate of \$6630 per month after the 6th month of release.
$\frac{{ - 400{t^2} - 2400t + 8000}}{{{{\left( {{t^2} + 20} \right)}^2}}} = \frac{0}{1}$
$- 400{t^2} - 2400t + 8000 = 0$
$- 400({t^2} + 6t - 20) = 0$
${t^2} + 6t - 20 = 0$
$t = \frac{{ - 6 \pm \sqrt {{6^2} - 4(1)( - 20)} }}{{2(1)}} = \frac{{ - 6 \pm \sqrt {36 + 80} }}{2} = \frac{{ - 6 \pm \sqrt {116} }}{2} = \frac{{ - 6 \pm 10.77}}{2}$
The profit begins to decrease 2.385 months after the release of the album.
Comcast has determined a model that predicts the total number of subscribers it will have when it expands to a new geographic region. The total number of subscribers $S$ (in thousands) $t$ months after the expansion is given by $S(t) = \frac{{150t}}{{t + 6}}$
$S'(t) = \frac{{150(t + 6) - 1(150t)}}{{{{\left( {t + 6} \right)}^2}}}$
$S'(t) = \frac{{150t + 900 - 150t}}{{{{\left( {t + 6} \right)}^2}}}$
$S'(t) = \frac{{900}}{{{{\left( {t + 6} \right)}^2}}}$
$S(15) = \frac{{150(15)}}{{15 + 6}} = 107.14$
$S'(15) = \frac{{900}}{{{{\left( {15 + 6} \right)}^2}}} = 2.04$
The total number of subscribers after 15 months is 107,140.The number of subscribers is increasing at a rate of about 2040 per month after the 15th month.
$S(16) \approx 107.14 + 2.04 \approx 109.18$
At 16 months, the number of subscribers will be approximately 109,180.
Find the derivative of each of the following.
$p(t) = {({t^3} + 4)^5}$
$p'(t) = 5{\left( {{t^3} + 4} \right)^4}(3{t^2})$
$p'(t) = 15{t^2}{\left( {{t^3} + 4} \right)^4}$
$f(x) = \sqrt {{x^2} - 144} $
$ f(x)= {({x^2} - 144)^{1/2}}$
$f'(x) = \frac{1}{2}{\left( {{x^2} - 144} \right)^{ - 1/2}}(2x)$
$f'(x) = \frac{{2x}}{{2\sqrt {{x^2} - 144} }}$
$f'(x) = \frac{x}{{\sqrt {{x^2} - 144} }}$
$g(x) = {e^{2{x^2} - 5x + 4}}$
$g'(x) = {e^{2{x^2} - 5x + 4}}(4x - 5)$
$p(a) = \ln \;({a^4} + 4a)$
$p'(a) = \frac{1}{{{a^4} + 4a}}(4{a^3} + 4)$
$p'(a) = \frac{{4{a^3} + 4}}{{{a^4} + 4a}}$
$y = \frac{1}{{\sqrt[3]{{x - {x^3}}}}}$
$y={{\left( x-{{x}^{3}} \right)}^{{}^{-1}/{}_{3}}}$
$y'=-\frac{1}{3}{{\left( x-{{x}^{3}} \right)}^{-4/3}}(1-3{{x}^{2}})$
$y'=\frac{-(1-3{{x}^{2}})}{3{{\left( x-{{x}^{3}} \right)}^{4/3}}}$
$g(x) = 9{(2{x^2} + x - 7)^{ - 3}}$
$g'(x)=-27{{\left( 2{{x}^{2}}+x-7 \right)}^{-4}}(4x+1)$
$g'(x)=\frac{-27(4x+1)}{{{\left( 2{{x}^{2}}+x-7 \right)}^{4}}}$
$y = {(7 - 5\ln \;x)^3}$
$y'=3\left(7-5\ln x\right)^2\cdot\left(\frac{-5}x\right)$
$y'=\frac{-15\left(7-5lnx\right)^2}x$
$q(t) = \ln (\ln \;5t)$
$q'\left(t\right)=\frac1{\ln(5t)}\cdot\frac1{5t}\cdot5$
$q'(t) = \frac{1}{{t\ln 5t}}$
Find the equation for the tangent line to the curve $y = {\sqrt {{e^x} + 8}}$ at the point where $x = 0.$
when $x = 0$
$y = \sqrt {{e^0} + 8} = \sqrt {1 + 8} = \sqrt 9 = 3$
$(0,3)$
$y = {\left( {{e^x} + 8} \right)^{ 1/2}}$
$y' = \frac{1}{2}{\left( {{e^x} + 8} \right)^{ - 1/2}}({e^x})$
$y' = \frac{{{e^x}}}{{2\sqrt {{e^x} + 8} }}$
${m_{\tan }}= \frac{{{e^0}}}{{2\sqrt {{e^0} + 8} }} = \frac{1}{{2\sqrt 9 }} = \frac{1}{6}$
$y - 3 = \frac{1}{6}(x - 0)$
$y - 3 = \frac{1}{6}x$
$y = \frac{1}{6}x + 3$
The concentration of toxic material in a lake is related to the number of months that an manufacturing plant has been operating near the lake. This concentration of toxic material can be modeled by $A(t)={{(0.7{{t}^{1/4}}+5)}^{3}}$ where A is measured in parts per million (ppm).
$A'(t)=3\left(0.7t^{1/4}+5\right)^2\left(0.7\cdot\frac14\cdot t^{-3/4}\right)$
$A'(t)=\frac{2.1\left(0.7t^{1/4}+5\right)^2}{4t^{3/4}}$
$A'(t)=\frac{21\left(0.7t^{1/4}+5\right)^2}{40t^{3/4}}$
$A(20)={{(0.7{{(20)}^{1/4}}+5)}^{3}} = 272.14$
The concentration of toxic material after 20 months is $272.14$ ppm
$A'(20)=\frac{21\left(0.7(20)^{1/4}+5\right)^2}{40(20)^{3/4}}$
The concentration of toxic material after 20 months is increasing at a rate of $2.33$ ppm per month
$272.14+2.33 = 274.47$
The total estimated amount of toxic material in the lake at 21 months is $274.47$ ppm.
Find the derivative of each of the following.
$p(t) = {t^2}{(5t + 1)^3}$
$f={{t}^{2}}$
$f'=2t$
$g={{\left( 5t+1 \right)}^{3}}$
$g'=3{{\left( 5t+1 \right)}^{2}}(5)$
$g'=15{{\left( 5t+1 \right)}^{2}}$
$p'(t) = 2t{\left( {5t + 1} \right)^3} + {t^2}(15{\left( {5t + 1} \right)^2})$
$p'(t) = 2t{\left( {5t + 1} \right)^3} + 15{t^2}{\left( {5t + 1} \right)^2}$
$p'(t) = {\left( {5t + 1} \right)^2}\left[ {2t(5t + 1) + 15{t^2}} \right]$
$p'(t) = {\left( {5t + 1} \right)^2}\left[ {10{t^2} + 2t + 15{t^2}} \right]$
$p'(t) = {\left( {5t + 1} \right)^2}(25{t^2} + 2t)$
$h(x) = \frac{{\sqrt {{x^2} - 36} }}{x}$
$f = {\left( {{x^2} - 36} \right)^{1/2}}$
$f' = \frac{1}{2}{\left( {{x^2} - 36} \right)^{ - 1/2}}(2x)$
$f' = \frac{{2x}}{{2\sqrt {{x^2} - 36} }}$
$f' = \frac{x}{{\sqrt {{x^2} - 36} }}$
$g = x$
$g' = 1$
$h'(x) = \frac{{\frac{x}{{\sqrt {{x^2} - 36} }}(x) - 1\sqrt {{x^2} - 36} }}{{{x^2}}}$
$h'(x) = \frac{{\frac{{{x^2}}}{{\sqrt {{x^2} - 36} }} - \sqrt {{x^2} - 36} }}{{{x^2}}}$
$h'(x) = \frac{{{x^2} - ({x^2} - 36)}}{{{x^2}\sqrt {{x^2} - 36} }}$
$h'(x) = \frac{{36}}{{{x^2}\sqrt {{x^2} - 36} }}$
$r(x) = (2{x^2} - 3){(7x + 4)^3}$
$f = 2{x^2} - 3$
$f' = 4x$
$g = {\left( {7x + 4} \right)^3}$
$g' = 3{\left( {7x + 4} \right)^2}(7)$
$g' = 21{\left( {7x + 4} \right)^2}$
$r'(x) = 4x{\left( {7x + 4} \right)^3} + (2{x^2} - 3)\left[ {21{{\left( {7x + 4} \right)}^2}} \right]$
$r'(x) = {\left( {7x + 4} \right)^2}\left[ {4x(7x + 4) + 21(2{x^2} - 3)} \right]$
$r'(x) = {\left( {7x + 4} \right)^2}\left[ {28{x^2} + 16x + 42{x^2} - 63} \right]$
$r'(x) = {\left( {7x + 4} \right)^2}\left[ {70{x^2} + 16x - 63} \right]$
$p(a) = {a^3}(\ln \;({a^5} + 2a))$
$f = {a^3}$
$f' = 3{a^2}$
$g = \ln ({a^5} + 2a)$
$g' = \frac{1}{{{a^5} + 2a}} \cdot (5{a^4} + 2)$
$g' = \frac{{(5{a^4} + 2)}}{{{a^5} + 2a}}$
$p'(a) = 3{a^2}\ln ({a^5} + 2a) + {a^3}(\frac{{5{a^4} + 2}}{{{a^5} + 2a}})$
$p'(a) = 3{a^2}\ln ({a^5} + 2a) + \frac{{{a^3}(5{a^4} + 2)}}{{a({a^4} + 2)}}$
$p'(a) = 3{a^2}\ln ({a^5} + 2a) + \frac{{{a^2}(5{a^4} + 2)}}{{{a^4} + 2}}$
$y = \frac{{{e^{{x^2} + x}}}}{{\sqrt[{}]{{2x + 1}}}}$
$f = {e^{{x^2} + x}}$
$f' = {e^{{x^2} + x}}(2x + 1)$
$g = {\left( {2x + 1} \right)^{1/2}}$
$g' = \frac{1}{2}{\left( {2x + 1} \right)^{ - 1/2}}(2)$
$g' = \frac{2}{{2\sqrt {2x + 1} }}=\frac{1}{{\sqrt {2x + 1} }}$
$y'=\frac{(2x+1){{e}^{{{x}^{2}}+x}}{{\left( 2x+1 \right)}^{1/2}}-{{e}^{{{x}^{2}}+x}}{{\left( 2x+1 \right)}^{-1/2}}}{{{\left[ {{\left( 2x+1 \right)}^{1/2}} \right]}^{2}}}$
$y'=\frac{{{\left( 2x+1 \right)}^{3/2}}{{e}^{{{x}^{2}}+x}}-{{\left( 2x+1 \right)}^{-1/2}}{{e}^{{{x}^{2}}+x}}}{2x+1}$
$y'=\frac{{{\left( 2x+1 \right)}^{4/2}}{{e}^{{{x}^{2}}+x}}-{{e}^{{{x}^{2}}+x}}}{{{\left( 2x+1 \right)}^{3/2}}}$
$y'=\frac{{{e}^{{{x}^{2}}+x}}\left( {{\left( 2x+1 \right)}^{2}}-1 \right)}{{{\left( 2x+1 \right)}^{3/2}}}$
$h(x) = (6x + 5){({x^2} + 4x + 8)^{ - 2}}$
$f = 6x + 5$
$f' = 6$
$g = {\left( {{x^2} + 4x + 8} \right)^{ - 2}}$
$g' = - 2{\left( {{x^2} + 4x + 8} \right)^{ - 3}}(2x + 4)$
$h'(x) = 6{\left( {{x^2} + 4x + 8} \right)^{ - 2}} + (6x + 5)[ - 2{\left( {{x^2} + 4x + 8} \right)^{ - 3}}(2x + 4)]$
$h'(x) = 6{\left( {{x^2} + 4x + 8} \right)^{ - 2}} - 2(6x + 5)(2x + 4){\left( {{x^2} + 4x + 8} \right)^{ - 3}}$
$h'(x) = \frac{6}{{{{\left( {{x^2} + 4x + 8} \right)}^2}}} - \frac{{2(6x + 5)(2x + 4)}}{{{{\left( {{x^2} + 4x + 8} \right)}^3}}}$
$y = \frac{{\ln \;\sqrt[3]{{9x + 2}}}}{x}$
$f=\ln {{\left( 9x+2 \right)}^{{}^{1}/{}_{3}}}$
$f'=\frac1{\left(9x+2\right)^{1/3}}\cdot\frac13\left(9x+2\right)^{-2/3}\cdot\left(9\right)$
$f'=\frac9{3\left(9x+2\right)}$
$f'=\frac3{9x+2}$
$g=x$
$g'=1$
$y'=\frac{\frac{3}{9x+2}(x)-1\ln {{\left( 9x+2 \right)}^{{}^{1}/{}_{3}}}}{{{x}^{2}}}$
$y'=\frac{\frac{3x}{9x+2}-\ln {{\left( 9x+2 \right)}^{{}^{1}/{}_{3}}}}{{{x}^{2}}}$
$y'=\frac{3x-(9x+2)\ln {{\left( 9x+2 \right)}^{{}^{1}/{}_{3}}}}{{{x}^{2}}(9x+2)}$
$q(x) = \frac{{{e^{2x}}}}{{{e^{3x}} + {e^{4x}}}}$
$f = {e^{2x}}$
$f' = {e^{2x}}(2)$
$f' = 2{e^{2x}}$
$g = {e^{3x}} + {e^{4x}}$
$g' = {e^{3x}}(3) + {e^{4x}}(4)$
$g' = 3{e^{3x}} + 4{e^{4x}}$
$q'(x) = \frac{{2{e^{2x}}({e^{3x}} + {e^{4x}}) - {e^{2x}}(3{e^{3x}} + 4{e^{4x}})}}{{{{({e^{3x}} + {e^{4x}})}^2}}}$
$q'(x) = \frac{{2{e^{5x}} + 2{e^{6x}} - 3{e^{5x}} - 4{e^{6x}}}}{{{{({e^{3x}} + {e^{4x}})}^2}}}$
$q'(x) = \frac{{ - {e^{5x}} - 2{e^{6x}}}}{{{{({e^{3x}} + {e^{4x}})}^2}}}$
Find an equation for the tangent line to the curve $n(x) = {x^2}\ln\;x$ at the point where $x = e.$
$n(e) = {e^2}\ln e= {e^2}(1) = {e^2}$
$(e,{e^2})$
$f = {x^2}$
$f' = 2x$
$g = \ln x$
$g' = \frac{1}{x}$
$n'(x) = 2x\ln x + {x^2}\left( {\frac{1}{x}} \right)$
$n'(x) = 2x\ln x + x$
${m_{\tan }} = n'(e) = 2e\ln e + e$
${m_{\tan }} = 2e + e = 3e$
${m_{\tan }}=3e$
$y - {e^2} = 3e(x - e)$
$y - {e^2} = 3ex - 3{e^2}$
$y = 3ex - 3{e^2} + {e^2}$
$y = 3ex - 2{e^2}$
The number of people in Knoxville who contract the flu can be modeled by $P(t) = \frac{{15,000}}{{50{e^{ - 0.3t}} + 1}}$, where $P$ is the number of people who contract the flu and $t$ is the number of days after the outbreak began.
$f = 15000$
$f' = 0$
$g = 50{e^{ - 0.3t}} + 1$
$g' = 50{e^{ - 0.3t}}( - 0.3) + 0$
$g' = - 15{e^{ - 0.3t}}$
$P'(t) = \frac{{0(50{e^{ - 0.3t}} + 1) - 15000( - 15{e^{ - 0.3t}})}}{{{{\left( {50{e^{ - 0.3t}} + 1} \right)}^2}}}$
$P'(t) = \frac{{225000{e^{ - 0.3t}}}}{{{{\left( {50{e^{ - 0.3t}} + 1} \right)}^2}}}$
$P(4)=934$
$P’(4)=262.76$
After 4 days, 934 people have the flu and 263 more will get the flu the next day.
$934+263=1197$
On the fifth day, approximately 1197 people will have the flu.
Find the derivative of each function. Show all of your work and simplify your answer.
$h(x)=\left(\frac5{x^2}-3\right)\left(6x^2+1\right)$
$h(x)=\left(5x^{-2}-3\right)\left(6x^2+1\right)$
$f=5x^{-2}-3$
$f'=-10x^{-3}$
$g=6x^2+1$
$g'=12x$
$h'(x)=-10x^{-3}\left(6x^2+1\right)+12x\left(5x^{-2}-3\right) $
$h'(x)=-60x^{-1}-10x^{-3}+60x^{-1}-36x$
$h'(x)=\frac{-10}{x^3}-36x$
$y=\sqrt[5]{\left(7x-8\right)^3}$
$y=\left(7x-8\right)^{3/5}$
$y'=\frac35\left(7x-8\right)^{-2/5}\left(7\right) $
$y'=\frac{21}{5\left(7x-8\right)^{2/5}}$
$y'=\frac{21}{5\sqrt[5]{\left(7x-8\right)^2}}$
$k(n)=\frac{4n-7}{\left(2n^4+5\right)^2}$
$f=4n-7$
$f'=4$
$g=\left(2n^4+5\right)^2$
$g'=2\left(2n^4+5\right)^1\left(8n^3\right)$
$g'=16n^3\left(2n^4+5\right)$
$k'(n)=\frac{4\left(2n^4+5\right)^2-16n^3\left(2n^4+5\right)\left(4n-7\right)}{\left[\left(2n^4+5\right)^2\right]^2}$
$k'(n)=\frac{4\left(2n^4+5\right)^2-16n^3\left(2n^4+5\right)\left(4n-7\right)}{\left(2n^4+5\right)^4}$
$k'(n)=\frac{4\left(2n^4+5\right)-16n^3\left(4n-7\right)}{\left(2n^4+5\right)^3}$
$k'(n)=\frac{8n^4+20-64n^4+112n^3}{\left(2n^4+5\right)^3}$
$k'(n)=\frac{-56n^4+112n^3+20}{\left(2n^4+5\right)^3}$
$f(x)=\ln\left(5x^2+3x\right)$
$f'(x)=\frac1{5x^2+3x}\cdot\left(10x+3\right)$
$f'(x)=\frac{10x+3}{5x^2+3x}$
$r(w)=\left(w^2-2\right)e^{3w^2-7}$
$f=\left(w^2-2\right)$
$f'=2w$
$g=e^{3w^2-7}$
$g'=e^{3w^2-7}\left(6w\right)$
$g'=6w\cdot e^{3w^2-7}$
$r'\left(w\right)=2w\cdot e^{3w^2-7}+\left(w^2-2\right)\cdot6w\cdot e^{3w^2-7}$
$r'\left(w\right)=e^{3w^2-7}\left[2w+\left(w^2-2\right)\cdot6w\right] $
$r'\left(w\right)=e^{3w^2-7}\left[2w+6w^3-12w\right] $
$r'\left(w\right)=e^{3w^2-7}\left[6w^3-10w\right] $
$r'\left(w\right)=2we^{3w^2-7}\left[3w^2-5\right] $