Derivatives of Quotients
If $y = \;\frac{f(x)}{g(x)},$
then $y' = \;\frac{{f'(x)\;g(x)\; - \;f(x)\;g'(x)\;}}{{[{g(x)}}]^2}.$
Two Methods for Finding the Derivative:
Find the derivative two different ways.
$r\left( x \right) = \;\frac{{{x^5}+4}}{{{x^2}}}$
Simplifying and Using Power Rule
$r(x) = \frac{{{x^5}}}{{{x^2}}} + \frac{4}{{{x^2}}}$
$r(x) = {x^3} + 4{x^{ - 2}}$
${r'}(x) = 3{x^2} - 8{x^{ - 3}}$
${r'}(x) = 3{x^2} - \frac{8}{{{x^3}}}$
Using Quotient Rule
$f = {x^5} + 4$
$f' = 5{x^4}$
$g = {x^2}$
$g' = 2x$
$r'(x) = \frac{{5{x^4}({x^2}) - 2x({x^5} + 4)}}{{({x^2})^2}}$
$r'(x) = \frac{{5{x^6} - 2{x^6} - 8x}}{{{x^4}}}$
$r'(x) = \frac{{3{x^6} - 8x}}{{{x^4}}}$
$r'(x) = \frac{{x(3{x^5} - 8)}}{{{x^4}}}$
$r'(x) = \frac{{3{x^5} - 8}}{{{x^3}}}$
$r'(x) = 3{x^2} - \frac{8}{{{x^3}}}$ (which is the same answer as part a)
Find the Derivative of each Function using the Quotient Rule.
$b\left( x \right) = \;\frac{{4x}}{{3x + 8}}$
$f =4x$
$f' =4$
$g =3x+8$
$g' = 3$
$b'(x) = \frac{{4(3x + 8) - 3(4x)}}{{{{\left( {3x + 8} \right)}^2}}}$
$b'(x) = \frac{{12x + 32 - 12x}}{{{{\left( {3x + 8} \right)}^2}}}$
$b'(x) = \frac{{32}}{{{{\left( {3x + 8} \right)}^2}}}$
$c\left( x \right) = \;\frac{{{x^2}\; - 9}}{{{x^2}\; + 1}}$
$f ={x^2} - 9$
$f' =2x$
$g ={x^2} + 1$
$g' = 2x$
$c'(x) = \frac{{2x({x^2} + 1) - 2x({x^2} - 9)}}{{{{\left( {{x^2} + 1} \right)}^2}}}$
$c'(x) = \frac{{2{x^3} + 2x - 2{x^3} + 18x}}{{{{\left( {{x^2} + 1} \right)}^2}}}$
$c'(x) = \frac{{20x}}{{{{\left( {{x^2} + 1} \right)}^2}}}$
$h\left( x \right) = \;\frac{{1 + {e^x}}}{{1 - {e^x}}}$
$f =1 + {e^x}$
$f' = {e^x}$
$g = 1 - {e^x}$
$g' = - {e^x}$
$h'(x) = \frac{{{e^x}(1 - {e^x}) - \left( { - {e^x}} \right)(1 + {e^x})}}{{{{\left( {1 - {e^x}} \right)}^2}}}$
$h'(x) = \frac{{{e^x}(1 - {e^x}) + {e^x}(1 + {e^x})}}{{{{\left( {1 - {e^x}} \right)}^2}}}$
$h'(x) = \frac{{{e^x} - {e^{2x}} + {e^x} + {e^{2x}}}}{{{{\left( {1 - {e^x}} \right)}^2}}}$
$h'(x) = \frac{{2{e^x}}}{{{{\left( {1 - {e^x}} \right)}^2}}}$
$j\left( x \right) = \;\frac{{3x}}{{4 + \ln x}}$
$f =3x$
$f' = 3$
$g =4 + \ln x$
$g' = \frac{1}{x}$
$j'(x) = \frac{{3(4 + \ln x) - \frac{1}{x}(3x)}}{{{{\left( {4 + \ln x} \right)}^2}}}$
$j'(x) = \frac{{12 + 3\ln x - 3}}{{{{\left( {4 + \ln x} \right)}^2}}}$
$j'(x) = \frac{{9 + 3\ln x}}{{{{\left( {4 + \ln x} \right)}^2}}}$
Find $\frac{{dy}}{{dw}}$ for $y = \;\frac{{2{w^4}\; - \;{w^3}}}{{6w - 1}}$
$f =2{w^4} - {w^3}$
$f' = 8{w^3} - 3{w^2}$
$g = 6w - 1$
$g' = 6$
$\frac{{dy}}{{dw}} = \frac{{(8{w^3} - 3{w^2})(6w - 1) - 6(2{w^4} - {w^3})}}{{{{\left( {6w - 1} \right)}^2}}}$
$\frac{{dy}}{{dw}} = \frac{{48{w^4} - 8{w^3} - 18{w^3} + 3{w^2} - 12{w^4} + 6{w^3}}}{{{{\left( {6w - 1} \right)}^2}}}$
$\frac{{dy}}{{dw}} = \frac{{36{w^4} - 20{w^3} + 3{w^2}}}{{{{\left( {6w - 1} \right)}^2}}}$
$h\left( x \right) = \;\frac{{3x - 7}}{{2x - 1}}$
$f =3x - 7$
$f' = 3$
$g = 2x - 1$
$g' = 2$
$h'(x) = \frac{{3(2x - 1) - 2(3x - 7)}}{{{{\left( {2x - 1} \right)}^2}}}$
$h'(x) = \frac{{6x - 3 - 6x + 14}}{{{{\left( {2x - 1} \right)}^2}}}$
$h'(x) = \frac{{11}}{{{{\left( {2x - 1} \right)}^2}}}$
Point $\quad h(2) = \frac{{3\cdot2 - 7}}{{2\cdot2 - 1}} = \frac{{ - 1}}{3}$
$(2, - \frac{1}{3})$
Slope $\quad {m_{tan}} = f'(2) = \frac{{11}}{{{{\left( {2\cdot2 - 1} \right)}^2}}} = \frac{{11}}{{{3^2}}} = \frac{{11}}{9}$
${m_{tan}} = \frac{{11}}{9}$
Equation of Line: $y + \frac{1}{3} = \frac{{11}}{9}(x - 2)$
$y + \frac{1}{3} = \frac{{11}}{9}x - \frac{{22}}{9}$
$y = \frac{{11}}{9}x - \frac{{25}}{9}$
$\frac{{11}}{{{{(2x - 1)}^2}}} = 0$
$\frac{{11}}{{{{(2x - 1)}^2}}} = \frac{0}{1}$
Cross Multiply
$0 \ne 11$ There is no value of x where $h'(x) =0.$
Derivatives with Radicals
Find y’ for $y = \;\frac{{6\sqrt[3]{x}}}{{2{x^2}\; - 5x + 1}}.$
$f(x)=\frac{{6{x^{1/3}}}}{{2{x^2} - 5x + 1}}$
$f =6{x^{1/3}}$
$f' = \frac{1}{3}\cdot 6{x^{ - 2/3}}$
$f' = 2{x^{ - 2/3}}$
$g = 2{x^2} - 5x + 1$
$g' = 4x - 5$
$y' = \frac{{2{x^{ - 2/3}}(2{x^2} - 5x + 1) - (4x - 5)(6{x^{1/3}})}}{{{{\left( {2{x^2} - 5x + 1} \right)}^2}}}$
$y' = \frac{{4{x^{4/3}} - 10{x^{1/3}} + 2{x^{ - 2/3}} - 24{x^{4/3}} + 30{x^{1/3}}}}{{{{\left( {2{x^2} - 5x + 1} \right)}^2}}}$
$y' = \frac{{ - 20{x^{4/3}} + 20{x^{1/3}} + 2{x^{ - 2/3}}}}{{{{\left( {2{x^2} - 5x + 1} \right)}^2}}}$
Multiply every term by ${{x}^{2/3}}.$
$y' = \frac{{ - 20{x^2} + 20x + 2}}{{\sqrt[3]{{{x^2}}}{{(2{x^2} - 5x + 1)}^2}}}$
Find $\frac{{dy}}{{dx}}$ for $y = \;\frac{{\;2{x^2}\; - 2x + 3\;}}{{\sqrt[4]{x}}}.$
$f =2{x^2} - 2x + 3$
$f' = 4x - 2$
$g = {x^{1/4}}$
$g' = \frac{1}{4}{x^{ - 3/4}}$
$\frac{{dy}}{{dx}} = \frac{{(4x - 2)({x^{1/4}}) - \frac{1}{4}{x^{ - 3/4}}(2{x^2} - 2x + 3)}}{{{{\left( {{x^{1/4}}} \right)}^2}}}$
$\frac{{dy}}{{dx}} = \frac{{4{x^{5/4}} - 2{x^{1/4}} - \frac{1}{2}{x^{5/4}} + \frac{1}{2}{x^{1/4}} - \frac{3}{4}{x^{ - 3/4}}}}{{{x^{1/2}}}}$
$\frac{{dy}}{{dx}} = \frac{{\left( {\frac{7}{2}{x^{5/4}} - \frac{3}{2}{x^{1/4}} - \frac{3}{4}{x^{ - 3/4}}} \right)}}{{({x^{1/2}})}}$
$\frac{{dy}}{{dx}} = \frac{{\left( {\frac{7}{2}{x^{5/4}} - \frac{3}{2}{x^{1/4}} - \frac{3}{4}{x^{ - 3/4}}} \right)\cdot 4}}{{({x^{1/2}})\cdot 4}}$
$\frac{{dy}}{{dx}} = \frac{{14{x^{5/4}} - 6{x^{1/4}} - 3{x^{ - 3/4}}}}{{(4{x^{1/2}})}}$
$\frac{dy}{dx}=\frac{14{{x}^{5/4}}-6{{x}^{1/4}}-\frac{3}{{{x}^{{}^{3}/{}_{4}}}}}{4{{x}^{1/2}}}$
Multiply every term by ${{x}^{3/4}}.$
$\frac{dy}{dx}=\frac{14{{x}^{8/4}}-6{{x}^{4/4}}-3}{(4{{x}^{1/2}}){{x}^{3/4}}}$
$\frac{dy}{dx}=\frac{14{{x}^{2}}-6x-3}{4{{x}^{5/4}}}$
Applications
A cable company has installed a new television system in a city. The total number N (in thousands) of subscribers t months after the installation of the system is given by $N\left( t \right) = \;\frac{{178t}}{{t + 5}}.$
$f =178t$
$f' = 178$
$g = t + 5$
$g' = 1$
$N'(t) = \frac{{178(t + 5) - 1(178t)}}{{{{\left( {t + 5} \right)}^2}}}$
$N'(t) = \frac{{178t + 890 - 178t}}{{{{\left( {t + 5} \right)}^2}}}$
$N'(t) = \frac{{890}}{{{{\left( {t + 5} \right)}^2}}}$
$N(12) = \frac{{178(12)}}{{12 + 5}} = \frac{{2136}}{{17}} = 125.647$
$N'(12) = \frac{{890}}{{{{\left( {12 + 5} \right)}^2}}} = \frac{{890}}{{{{17}^2}}} = \frac{{890}}{{289}} = 3.0796$
At 12 months the cable company has 125,647 subscribers and that number is increasing at a rate of 3080 subscribers per month.
$N(13)\approx N(12)+N'(12)$
125,647 + 3080 = 128,727
Based on the number of subscribers after 12 months, there will be approximately 128,727 subscribers after 13 months.
According to economic theory, the supply x of a quantity in a free market increases as the price p increases. Suppose the number x of baseball gloves a retail chain is willing to sell per week at a price of \$p is given by $$x = \;\frac{{100p}}{{0.1p + 1}} \quad\ 30.00 \le p \le 190.00.$$
$f =100p$
$f' = 100$
$g =0.1p + 1$
$g' = 0.1$
$\frac{{dx}}{{dp}} = \frac{{100(0.1p + 1) - 0.1(100p)}}{{{{\left( {0.1p + 1} \right)}^2}}}$
$\frac{{dx}}{{dp}} = \frac{{10p + 100 - 10p}}{{{{\left( {0.1p + 1} \right)}^2}}}$
$\frac{{dx}}{{dp}} = \frac{{100}}{{{{\left( {0.1p + 1} \right)}^2}}}$
Supply when $p=\$40\quad\quad\quad x(40) = \frac{{100(40)}}{{0.1(40) + 1}} = \frac{{4000}}{5} = 800\;gloves$
IRC when $p=\$40\quad\quad\quad \frac{{dx}}{{dp}}(40)=\frac{{100}}{{{{\left({0.1\cdot40+1}\right)}^2}}} = \frac{{100}}{{25}} = 4\;gloves$
When the price of the baseball gloves is $40, the retail chain can sell 800 gloves per week, and the number of gloves sold is increasing at a rate of 4 gloves per week.
800+4=804
If the price increased to $41, 804 gloves would be sold each week.
Find the derivative of each of the following.
$h(x) = \frac{{6x + 5}}{{3x - 8}}$
$f = 6x + 5$
$f' = 6$
$g = 3x - 8$
$g' = 3$
$h'(x) = \frac{{6(3x - 8) - 3(6x + 5)}}{{{{\left( {3x - 8} \right)}^2}}}$
$h'(x) = \frac{{18x - 48 - 18x - 15}}{{{{\left( {3x - 8} \right)}^2}}}$
$h'(x) = \frac{{ - 63}}{{{{\left( {3x - 8} \right)}^2}}}$
$y = \frac{{{x^3} - 5x}}{{4{x^2}}}$
$f = {x^3} - 5x$
$f' = 3{x^2} - 5$
$g = 4{x^2}$
$g' = 8x$
$y' = \frac{{(3{x^2} - 5)(4{x^2}) - 8x({x^3} - 5x)}}{{{{\left( {4{x^2}} \right)}^2}}}$
$y' = \frac{{12{x^4} - 20{x^2} - 8{x^4} + 40{x^2}}}{{{{\left( {4{x^2}} \right)}^2}}}$
$y' = \frac{{4{x^4} + 20{x^2}}}{{16{x^4}}}$
$y' = \frac{{4{x^2}({x^2} + 5)}}{{16{x^4}}}$
$y' = \frac{{{x^2} + 5}}{{4{x^2}}}$
$r(x) = \frac{2e^x}{x-7}$
$f = 2e^x$
$f' = 2e^x$
$g = x-7$
$g' = 1$
$r'(x) = \frac{2e^x(x-7)-2e^x\left(1\right)}{\left(x-7\right)^2}$
$r'(x) = \frac{2xe^x-14e^x-2e^x}{\left(x-7\right)^2}$
$r'(x) = \frac{2xe^x-16e^x}{\left(x-7\right)^2}$
$h(x) = \frac{x+7}{-3e^x}$
$f = x+7$
$f' = 1$
$g = -3e^x$
$g' = -3e^x$
$h'(x) = \frac{1(-3e^x)-\left(-3e^x\right)(x+7)}{\left(-3e^x\right)^2}$
$h'(x) = \frac{-3e^x\left[1-(x+7)\right]}{\left(-3e^x\right)^2}$
$h'(x) = \frac{\left[1-(x+7)\right]}{-3e^x}$
$h'(x) = \frac{1-x-7}{-3e^x}$
$h'(x) = \frac{-x-6}{-3e^x}$
$h'(x) = \frac{x+6}{3e^x}$
$p(x) = \frac{\sqrt x}{2x-4}$
$f = x^\frac12$
$f' = \frac12x^{-\frac12}$
$g = 2x-4$
$g' = 2$
$p'(x) = \frac{\displaystyle\frac12x^\frac12\left(2x-4\right)-2\left(x^\frac12\right)}{\left(2x-4\right)^2}$
$p'(x) = \frac{\displaystyle x^\frac12-2x^\frac{-1}2-2x^\frac12}{\left(2x-4\right)^2}$
$p'(x) =\frac{\displaystyle-1x^\frac12-2x^\frac{-1}2}{\left(2x-4\right)^2}$
Multiply all terms by $x^\frac12.$
$p'(x) = \frac{\displaystyle-x-2}{\sqrt x\left(2x-4\right)^2}$
$r(x) = \frac{\sqrt[3]x}{3x-9}$
$f = x^\frac13$
$f' = \frac13x^{-\frac23}$
$g = 3x-9$
$g' = 3$
$r'(x) = \frac{\displaystyle\frac13x^\frac{-2}3\left(3x-9\right)-3(x^\frac13)}{\left(3x-9\right){\displaystyle{}^2}}$
$r'(x) = \frac{\displaystyle1x^\frac13-3x^\frac{-2}3-3x^\frac13}{\left(3x-9\right){\displaystyle{}^2}}$
$r'(x) =\frac{\displaystyle-2x^\frac13-3x^\frac{-2}3}{\left(3x-9\right){\displaystyle{}^2}}$
Multiply all terms by $x^\frac23.$
$r'(x) =\frac{\displaystyle-2x-3}{\sqrt[3]{x^2}\left(3x-9\right){\displaystyle{}^2}}$
$w(x) = \frac{\ln x}{x^3}$
$f =\ln x$
$f' = \frac1x$
$g = x^3$
$g' = 3x^2$
$w'(x) = \frac{\displaystyle\frac1x\left(x^3\right)-3x^2\ln x}{\left(x^3\right)^2}$
$w'(x) = \frac{x^2-3x^2\ln x}{x^6}$
$w'(x) =\frac{1-3\ln x}{x^4}$
Determine the slope of the tangent line to the curve $R(x) = \frac{{{x^2} - 2x - 8}}{{{x^2} - 9}}$ at the point where $x = 0$.
$f = {x^2} - 2x - 8$
$f' = 2x - 2$
$g = {x^2} - 9$
$g' = 2x$
$R'(x) = \frac{{(2x - 2)({x^2} - 9) - 2x({x^2} - 2x - 8)}}{{{{\left( {{x^2} - 9} \right)}^2}}}$
$R'(x) = \frac{{2{x^3} - 18x - 2{x^2} + 18 - 2{x^3} + 4{x^2} + 16x}}{{{{\left( {{x^2} - 9} \right)}^2}}}$
$R'(x) = \frac{{2{x^2} - 2x + 18}}{{{{\left( {{x^2} - 9} \right)}^2}}}$
${m_{\tan }}=R'(0) = \frac{{2{{(0)}^2} - 2(0) + 18}}{{{{\left( {{0^2} - 9} \right)}^2}}} = \frac{{18}}{{81}} = \frac{2}{9}$
${m_{\tan }} =\frac{2}{9}$
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 zeros of the equation are $t=2.385$ and $t=-8.385.$ Note: $t=-8.385$ is not in the domain.
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.
$S(16) \approx 107.14 + 2.04 \approx 109.18$
$S(16)\approx S(15)+S'(15)$
At 16 months, the number of subscribers will be approximately 109,180.
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.