Derivatives of Products
If $y=f(x)\cdot g(x)$,
then $y' = f'(x)\cdot g(x)\; + \;\;f(x)\cdot g'(x).$
Two Methods for Finding the Derivative:
Find the derivative two different ways.
$m\left( x \right) = 2{x^3}\;\left( {{x^5} - 2} \right)$
Simplifying and Using Power Rule
$m(x) = 2{x^8} - 4{x^3}$
${m'}(x) = 16{x^7} - 12{x^2}$
Using the Product Rule
$f = 2{x^3}$
$f' = 6{x^2}$
$g = {x^5} - 2$
$g' = 5{x^4}$
$m'(x) = 6{x^2}({x^5} - 2) + 2{x^3}(5{x^4})$
$m'(x) = 6{x^7} - 12{x^2} + 10{x^7}$
$m'(x) = 16{x^7} - 12{x^2}$
Find the derivative using the Product Rule.
$n(x)=7x^2\left(2x^3+5\right)$
$f = 7{x^2}$
$f' = 14x$
$g = 2{x^3} + 5$
$g' = 6{x^2}$
$n'(x) = 14x(2{x^3} + 5) + 7{x^2}(6{x^2})$
$n'(x) = 28{x^4} + 70x + 42{x^4}$
$n'(x) = 70{x^4} + 70x$
$h\left( x \right) = 4{x^3}\;{e^x}$
$f = 4{x^3}$
$f' =12{x^2}$
$g = {e^x}$
$g' = {e^x}$
$h'(x) = 12{x^2}({e^x}) + 4{x^3}({e^x})$
$h'(x) = {e^x}(12{x^2} + 4{x^3})$
$h'(x) = 4{x^2}{e^x}(3 + x)$
$s\left( x \right) = 2{x^5}\ln x$
$f = 2{x^5}$
$f' =10{x^4}$
$g = \ln x$
$g' = \frac{1}{x}$
$s'(x) = 10{x^4}\ln x + 2{x^5}\left( {\frac{1}{x}} \right)$
$s'(x) = 10{x^4}\ln x + 2{x^4}$
$s'(x) = 2{x^4}(5\ln x + 1)$
$v\left( x \right) = \left( {8x + 1} \right)\left( {3{x^2}\; - 7} \right)$
$f = 8x + 1$
$f' =8$
$g = 3{x^2} - 7$
$g' = 6x$
$v'(x) = 8(3{x^2} - 7) + 6x(8x + 1)$
$v'(x) = 24{x^2} - 56 + 48{x^2} + 6x$
$v'(x) = 72{x^2} + 6x - 56$
Tangent Lines
$r\left( x \right) = \left( {5 - 4x} \right)\left( {1 + 3x} \right)$
$f = 5 - 4x$
$f' = - 4$
$g = 1 + 3x$
$g' = 3$
$r'(x) = - 4(1 + 3x) + 3(5 - 4x)$
$r'(x) = - 4 - 12x + 15 - 12x$
$r'(x) = - 24x + 11$
Point $\quad r(2) = (5 - 4\cdot2)(1 + 3\cdot2) = ( - 3)(7) = -21$
$(2, - 21)$
Slope $\quad {m_{tan}}= {r'}(2) = - 24(2) + 11=-48+11=-37$
${m_{tan}}=-37$
Equation of the tangent line:
$y + 21 = - 37(x - 2)$
$y + 21 = - 37x + 74$
$y = - 37x + 53$
$ - 24x + 11 = 0$
$ - 24x = - 11$
$x = \frac{{11}}{{24}}$
Derivatives with Radicals
Find y’ for $y = \sqrt x\left(x^2+3x-1\right).$
$y=x^\frac12\left(x^2+3x-1\right)$
$f =x^\frac12$
$f' =\frac12x^\frac{-1}2$
$g = x^2 +3x -1$
$g' = 2x+3$
$y' = \frac12x^\frac{-1}2\left(x^2+3x-1\right)+x^\frac12\left(2x+3\right)$
$y' = \frac12x^\frac32+\frac32x^\frac12-\frac12x^\frac{-1}2+2x^\frac32+3x^\frac12$
$y' = \frac52x^\frac32+\frac92x^\frac12-\frac12x^\frac{-1}2$
Find $\frac{{dy}}{{dx}}$ for $y = \sqrt[3]x\left(x^6+x^3\right).$
$y=x^\frac13(x^6+x^3)$
$f =x^\frac13$
$f' = \frac13x^{-\frac23}$
$g = x^6+x^3$
$g' = 6x^5+3x^2$
$\frac{{dy}}{{dx}} = \frac13x^{-\frac23}\left(x^6+x^3\right)+x^\frac13\left(6x^5+3x^2\right)$
$\frac{{dy}}{{dx}} = \frac13x^\frac{16}3+\frac13x^\frac73+6x^\frac{16}3+3x^\frac73$
$\frac{dy}{dx}=\frac{19}3x^\frac{16}3+\frac{10}3x^\frac73$
Applications
Calculators are sold to students for 100 dollars each. Three hundred students are willing to buy them at that price. For every 5 dollar decrease in price, there are 30 more students willing to buy the calculator. The revenue function is given by the formula $ R(d)=(100-5d)(300+30d) $.
Find $R'(d).$
$f = 100-5d$
$f' = - 5$
$g = 300+30d$
$g' = 30$
$ R'(d)=-5(300+30d)+30(100-5d) $
$ R'(d)=-1500-150d+3000-150d$
$ R'(d)=1500-300d$
$R(3) = (100-5(3))(300+30(3))=(85)(390) = 33150$
$R'(3) = 1500-300(3)=600$
At three \$5 reductions in price, the revenue is \$33,150. At that point the revenue is increasing at a rate of \$600 per $5 decrease in price.
$R(4)\approx R(3)+R'(3)$
33150 + 600 = 33750
The revenue will be approximately \$33,750 after four $5 price reductions.
Find the derivative of each of the following.
$p(x) = (6x-1)(5x+2)$
$f = 6x-1$
$f' = 6$
$g = 5x+2$
$g' = 5$
$p'(x) = 6(5x+2)+5(6x-1)$
$p'(x) = 30x+12+30x-5$
$p'(x) = 60x+7$
$p(t) = 2{t^2}({t^3} + 4t)$
$f = 2{t^2}$
$f' = 4t$
$g = {t^3} + 4t$
$g' = 3{t^2} + 4$
$p'(t) = 4t({t^3} + 4t) + 2{t^2}(3{t^2} + 4)$
$p'(t) = 4{t^4} + 16{t^2} + 6{t^4} + 8{t^2}$
$p'(t) = 10t{}^4 + 24{t^2}$
$h(z) = 3z^2e^z$
$f = 3z^2$
$f' = 6z$
$g = e^z$
$g' = e^z$
$h'(z) = 6z(e^z)+e^z(3z^2)$
$h'(z) = 6ze^z+3z^2e^z$
$h'(z) = 3ze^z(2+z)$
$p(x) = 2{e^x}({x^2} - 3x + 5)$
$f = 2{e^x}$
$f' = 2{e^x}$
$g = {x^2} - 3x + 5$
$g' = 2x - 3$
$p'(x) = 2{e^x}({x^2} - 3x + 5) + 2{e^x}(2x - 3)$
$p'(x) = 2{e^x}({x^2} - 3x + 5 + 2x - 3)$
$p'(x) = 2{e^x}({x^2} - x + 2)$
$y = -x^2\ln x$
$f = -x^2$
$f' = -2x$
$g = lnx$
$g' = \frac1x$
$y'(x) = -2xlnx+(\frac1x)(-x^2)$
$y'(x) = -2xlnx+\frac{-x^2}x$
$y'(x) = -2xlnx-x$
$r(x) = (x^3+x)\ln x$
$f = x^3+x$
$f' = 3x^2+1$
$g = lnx$
$g' = \frac1x$
$r'(x) = (3x^2+1)lnx+(\frac1x)(x^3+x)$
$r'(x) = (3x^2+1)lnx+\frac{x^3+x}x$
$r'(x) = (3x^2+1)lnx+x^2+1$
$k(x) = (2x - 5)({x^2} + 1)$
$f = 2x-5$
$f' = 2$
$g = x^2+1$
$g' = 2x$
$k'(x) = 2({x^2} + 1) + 2x(2x - 5)$
$k'(x) = 2{x^2} + 2 + 4{x^2} - 10x$
$k'(x) = 6{x^2} - 10x + 2$
$p(a) = ({a^2} - 2a + 7)(2{a^2} - a + 1)$
$f = {a^2} - 2a + 7$
$f' = 2a - 2$
$g = 2{a^2} - a + 1$
$g' = 4a - 1$
$p'(a) = (2a - 2)(2{a^2} - a + 1) + (4a - 1)({a^2} - 2a + 7)$
$p'(a) = 4{a^3} - 2{a^2} + 2a - 4{a^2} + 2a - 2 + 4{a^3} - 8{a^2} + 28a - {a^2} + 2a - 7$
$p'(a) = 8{a^3} - 15{a^2} + 34a - 9$
$p(t) = \sqrt t \;({t^2} + 5t - 2)$
$f = {t^{1/2}}$
$f' = \frac{1}{2}{t^{ - 1/2}}$
$g = {t^2} + 5t - 2$
$g' = 2t + 5$
$p'(t)=\frac{1}{2}{{t}^{-1/2}}({{t}^{2}}+5t-2)+{{t}^{1/2}}(2t+5)$
$p'(t)=\frac{1}{2}{{t}^{3/2}}+\frac{5}{2}{{t}^{1/2}}-{{t}^{-1/2}}+2{{t}^{3/2}}+5{{t}^{1/2}}$
$p'(t)=\frac{5}{2}{{t}^{3/2}}+\frac{15}{2}{{t}^{1/2}}-{{t}^{-1/2}}$
$p'(t)=\frac{5}{2}{{t}^{3/2}}+\frac{15}{2}{{t}^{1/2}}-{\frac{1}{{t}^{1/2}}}$
$q(x) = (\sqrt x+1)(\sqrt x-3)$
$f = x^{1/2}+1$
$f' = \frac{1}{2}{x^{ - 1/2}}$
$g = x^{1/2}-3$
$g' = \frac{1}{2}{x^{ - 1/2}}$
$q'(x)=\frac12x^\frac{-1}2\left(x^\frac12-3\right)+\frac12x^\frac{-1}2\left(x^\frac12+1\right)$
$q'(x)=\frac12-\frac32x^\frac{-1}2+\frac12+\frac12x^\frac{-1}2$
$q'(x)=1-1x^\frac{-1}2$
$q'(x)=1-\frac1{\sqrt x}$
$y = x^{-4}(x^2-7x+1)$
$f = x^{-4}$
$f' = -4x^{-5}$
$g = x^2-7x+1$
$g' = 2x-7$
$y' = -4x^{-5}(x^2-7x+1)+(2x-7)(x^{-4})$
$y' = -4x^{-3}+28x^{-4}-4x^{-5}+2x^{-3}-7x^{-4}$
$y' = -2x^{-3}+21x^{-4}-4x^{-5}$
$y' = x^{-3}(-2+21x^{-1}-4x^{-2})$
$y = \frac2{x^2}\left(3x^4-6x+2\right)$
$f = 2x^{-2}$
$f' =-4x^{-3}$
$g = 3x^4-6x+2$
$g' = 12x^3-6$
$y' = -4x^{-3}(3x^4-6x+2)+2x^{-2}(12x^3-6)$
$y' = -12x+24x^{-2}-8x^{-3}+24x-12x^{-2}$
$y' = -8x^{-3}+12x^{-2}+12x$
If $z(x) = x^2(x^3+1)$, find the slope of that tangent line at $x=1.$
Point: $z(1)=(1)^2(1^3+1)=(1)(1+1)=1(2)=2$
$(1,2)$
Slope:
$f =x^2$
$f' =2x$
$g = x^3+1$
$g' = 3x^2$
$z' = 2x(x^3+1)+3x^2(x^2)$
$z' = 2x^4+2x+3x^4$
$z' = 5x^4+2x$
$z'(1); = 5(1)^4+2(1)=7$
$m_{tan}=7$
Equation of the Tangent Line at $x=1$
$y-y_1=m(x-x_1)$
$y-2=7(x-1)$
$y-2=7x-7$
$y=7x-5$
The manager of a miniature golf course is planning to raise the ticket price per game. At the current price of \$6.50, an average of 81 games is played each day. The manager’s research suggests that for every \$0.50 increase in price, an average of four fewer games will be played each day.
$R(n) =$ (price)(number sold)
$R(n) = (6.50 + .50n)(81 - 4n)$
$R(n) = 526.50 + 14.5n - 2{n^2}$
$R'(n) = 14.5 - 4n$
If the price is increased by \$1.50, then there have been 3 fifty cent increases, so $n=3$.
$R'(3) = 2.5$
When the manager increases the price per ticket by \$$1.50$, the number of rounds played each day decreases but the actual revenue increases by \$$2.50$ per day. This provides the owner with an increase in revenue even though there is a decrease in the number of rounds played each day.
In 2012, the population of a city was 939,200, and the population was increasing at roughly 9500 per year. The average annual income was \$35,450 per capita, and this average was increasing at about $1400 per year. The function below models the total personal income for the population of the city.
$I(x)=(939,200+9500x)(35,450+1400x)$
Estimate the rate at which total personal income of the city's population will be rising in the year 2020.
$f =939200+9500x$
$f' =9500$
$g = 35450+1400x$
$g' = 1400$
$I'(x) = 9500(35450+1400x)+1400(939200+9500x)$
$I'(x) =336,775,000+13,300,000x+1,314,880,000+13,300,000x$
$I'(x) = 26,600,000x+1,651,655,000$
$I'(8); = 26,600,000(8)+1,651,655,000=1,864,455,000$
In 2020 the total personal income for the area is rising at a rate of $1,864,445,000 per year.