$ \frac d{dx}\left[f(x)\right]^n=n\left[f(x)\right]^{n-1}\cdot f'(x)$
$\frac d{dx}\ln\left[f(x)\right]=\frac1{f(x)}\cdot f'(x)$
$\frac d{dx}e^{f(x)}=e^{f(x)}\cdot f'\left(x\right)$
Find the derivative.
$f\left( x \right) = \;{\left( {8{x^2}\; - 7} \right)^5}$
$f'(x) = 5{\left( {8{x^2} - 7} \right)^4}(16x)$
$f'(x) = 80x{\left( {8{x^2} - 7} \right)^4}$
$f\left( x \right) = \;{e^{3{x^2}\; + 2x + 5}}$
$f'(x) = {e^{3{x^2} + 2x + 5}}(6x + 2)$
$f\left( x \right) = 2\ln \left( {9{x^2}\; - 5x + 21} \right)$
$f'(x) = \frac{2}{{9{x^2} - 5x + 21}}\cdot(18x - 5)$
$f'(x) = \frac{{36x - 10}}{{9{x^2} - 5x + 21}}$
Find the derivative
$f\left( x \right) = \;{\left( {4x - 5\ln x} \right)^7}$
$f'(x) = 7{\left( {4x - 5\ln x} \right)^6}(4 - \frac{5}{x})$
$f'(x) = 7(4 - \frac{5}{x}){\left( {4x - 5\ln x} \right)^6}$
Find the Horizontal Tangents
Find $f'(x)$ and simplify. Then find the equation of the tangent line to the graph of $f(x)$ at the given value of $x$. Find the values of $x$ where the tangent line is horizontal.
$f\left( x \right) = \;{\left( {3x + 13} \right)^{1/2}}\quad \quad$ at $x = 4$
$f'(x) = \frac{1}{2}{\left( {3x + 13} \right)^{ - 1/2}}(3)$
$f'(x) = \frac{3}{{2\sqrt {3x + 13} }}$
Slope: $\quad {m_{tan}} = f'(4) = \frac{3}{{2\sqrt {3\cdot4 + 13} }}$
$\quad\quad{m_{tan}} = \frac{3}{{2\sqrt {12 + 13} }}$
$\quad\quad{m_{tan}} = \frac{3}{{2\sqrt {25} }}$
$\quad\quad{m_{tan}} = \frac{3}{{2\cdot5}} = \frac{3}{{10}}$
Point: $f(4) = {\left( {3\cdot 4 + 13} \right)^{1/2}}$
$\quad\quad f(4) = {\left( {12 + 13} \right)^{1/2}}$
$\quad\quad f(4) = \sqrt {25} = 5$
$\quad\quad (4,5)$
$y - 5 = \frac{3}{{10}}(x - 4)$
$y - 5 = \frac{3}{{10}}x - \frac{{12}}{{10}}$
$y = \frac{3}{{10}}x - \frac{{6}}{{5}} + 5$
$y = \frac{3}{{10}}x + \frac{{19}}{5}$
Horizontal Tangent
Horizontal Tangents are found where $f'(x) = 0.$
$\frac{3}{{2\sqrt {3x + 13} }} = 0$
$\frac{3}{{2\sqrt {3x + 13} }} = \frac{0}{1}$
Cross multiply.
$3 \ne 0$
No Horizontal Tangents: There are no values of x where $f'(x) = 0.$
$f\left( x \right) = \;3{e^{2{x^2}\; + 5x - 4}}$
$f'(x) = 3{e^{2{x^2} + 5x - 4}}(4x + 5)$
$f'(x) = 3(4x + 5){e^{2{x^2} + 5x - 4}}$
$f\left( x \right) = \ln \left( {1 - {x^2} + 2{x^4}} \right)$
$f'(x) = \frac{1}{{1 - {x^2} + 2{x^4}}}\cdot( - 2x + 8{x^3})$
$f'(x) = \frac{{ - 2x + 8{x^3}}}{{1 - {x^2} + 2{x^4}}}$
$\frac{d}{{dt}}\;3{\left( {2{t^4}\; + {t^2}\;} \right)^{ - 5}}$
$\frac{d}{{dt}} = - 15{\left( {2{t^4} + {t^2}} \right)^{ - 6}}(8{t^3} + 2t)$
$ \frac{d}{{dt}} =\frac{{ - 120{t^3} - 30t}}{{{{\left( {2{t^4} + {t^2}} \right)}^6}}}$
$\frac{{dh}}{{dw\;}}\quad $ if $\quad h\left( w \right) = \;\sqrt[5]{{8w - 1}}$
$ h(w)= {\left( {8w - 1} \right)^{1/5}}$
$\frac{{dh}}{{dw}} = \frac{1}{5}{\left( {8w - 1} \right)^{ - 4/5}}(8) $
$ \frac{{dh}}{{dw}} =\frac{8}{{5\,\,\sqrt[5]{{{{\left( {8w - 1} \right)}^4}}}}}$
$G'\left( t \right)$ if $G\left( t \right) = \;{\left( {t - {e^{9t}}} \right)^2}$
$G'(t) = 2{\left( {t - {e^{9t}}} \right)^1}(1 - 9{e^{9t}})$
$ G'(t) = 2(t - {e^{9t}})(1 - 9{e^{9t}})$
$y'$ if $y=\ln{(x^2+7)}^\frac45$
$y' = \frac{4}{5}{\left[ {\ln ({x^2} + 7)} \right]^{ - 1/5}}(\frac{1}{{{x^2} + 7}})(2x)$
$y' = \frac{{8x}}{{5({x^2} + 7){{\left( {\ln ({x^2} + 7)} \right)}^{1/5}}}}$
$\frac{d}{{dw}}\;\frac{1}{{{{\left( {{w^2}\; - 5} \right)}^3}}}$
$\frac{d}{{dw}} = \left(w^2-5\right)^{-3} $
$\frac{d}{{dw}} = - 3{\left( {{w^2} - 5} \right)^{ - 4}}(2w)$
$\frac{d}{{dw}} = \frac{{ - 6w}}{{{{\left( {{w^2} - 5} \right)}^4}}}$
Find $f'(x)$ and simplify.
$f\left( x \right) = {x^2}\;{\left( {3 - 2x} \right)^4}$ $h = {x^2}$ $h' = 2x$ $g = {\left( {3 - 2x} \right)^4}$ $g' = 4{\left( {3 - 2x} \right)^3}( - 2)=- 8{\left( {3 - 2x} \right)^3}$ $f'(x) = 2x{\left( {3 - 2x} \right)^4} + x{}^2( - 8{\left( {3 - 2x} \right)^3})$ $f'(x) = {\left( {3 - 2x} \right)^3}\left[ {2x(3 - 2x) - 8{x^2}} \right]$ $f'(x) = {\left( {3 - 2x} \right)^3}\left[ {6x - 4{x^2} - 8{x^2}} \right]$ $f'(x) = {\left( {3 - 2x} \right)^3}( - 12{x^2} + 6x)$
$f\left( x \right) = \;\frac{{{x^4}}}{{{{\left( {2x - 5} \right)}^2}}}$ Point: $\quad f(2) = \frac{{{2^4}}}{{{{\left( {2\cdot2 - 5} \right)}^2}}} = \frac{{16}}{{{{\left( { - 1} \right)}^2}}} = 16$ $(2,16)$ $h = {x^4}$ $h' = 4{x^3}$ $g ={\left( {2x - 5} \right)^2}$ $g' =2{\left( {2x - 5} \right)^1}(2)=4(2x - 5)$ $f'(x) = \frac{{4{x^3}{{\left( {2x - 5} \right)}^2} - {x^4}\cdot{4(2x - 5)}}}{{{{\left[ {{{\left( {2x - 5} \right)}^2}} \right]}^2}}}$ $f'(x) = \frac{{4{x^3}{{\left( {2x - 5} \right)}^2} - 4{x^4}(2x - 5)}}{{{{\left( {2x - 5} \right)}^4}}}$ $f'(x) = \frac{(2x-5)[{4x^3}(2x-5) - 4{x^4}]}{(2x - 5)^4}$ $f'(x) = \frac{{4{x^3}(2x - 5) - 4{x^4}}}{{{{\left( {2x - 5} \right)}^3}}}$ $f'(x) = \frac{{8{x^4} - 20{x^3} - 4{x^4}}}{{{{\left( {2x - 5} \right)}^3}}}$ $f'(x) = \frac{{4{x^4} - 20{x^3}}}{{{{\left( {2x - 5} \right)}^3}}}$
$h'\left( x \right)\quad $ if $\quad h\left( x \right) = \;\frac{{{e^{4x}}}}{{{x^3}\; + 9x}}$ $f = {e^{4x}}$ $f' = 4{e^{4x}}$ $g = {x^3} + 9x$ $g' = 3{x^2} + 9$ $h'(x) = \frac{{4{e^{4x}}({x^3} + 9x) - {e^{4x}}(3{x^2} + 9)}}{{{{\left( {{x^3} + 9x} \right)}^2}}}$ $h'(x) = \frac{{{e^{4x}}(4{x^3} + 36x - 3{x^2} - 9)}}{{{{\left( {{x^3} + 9x} \right)}^2}}}$ $h'(x) = \frac{{{e^{4x}}(4{x^3} - 3{x^2} + 36x - 9)}}{{{{\left( {{x^3} + 9x} \right)}^2}}}$
$\frac{d}{{dx}}\;\left[ {{x^5}\;\;ln\left( {3 + {x^5}\;} \right)} \right]$ $f = {x^5}$ $f' = 5{x^4}$ $g = \ln (3 + {x^5})$ $g' =\frac{1}{{3 + {x^5}}}\cdot5{x^4} = \frac{{5{x^4}}}{{3 + {x^5}}}$ $\frac{d}{{dx}} = 5{x^4}(\ln (3 + {x^5})) + \frac{{5{x^4}}}{{3 + {x^5}}}({x^5})$ $\frac{d}{{dx}} = 5{x^4}\ln (3 + {x^5}) + \frac{{5{x^9}}}{{3 + {x^5}}}$3.5B Lecture
3.5B Group Work
Horizontal Tangents
Find $f'(x)$ and simplify. Then find the equation of the tangent line to the graph of $f(x)$ at the given value of $x$. Find the values of $x$ where the tangent line is horizontal.
$f\left( x \right)={{e}^{\sqrt{x}}}\quad\quad$ when $x=1$
Point: $\quad f(1) = {e^{\sqrt 1 }} = {e^1} = e$
$(1,e)$
$f'(x)={e^{{x^{1/2}}}}$
$f'(x) = {e^{{x^{1/2}}}}\cdot\frac{1}{2}{x^{ - 1/2}}$
$f'(x) = {e^{\sqrt x }}\cdot\frac{1}{{2\sqrt x }}$
$f'(x) = \frac{{{e^{\sqrt x }}}}{{2\sqrt x }}$
Slope: $\quad {m_{tan}} = f'(1) = \frac{{{e^{\sqrt 1 }}}}{{2\sqrt 1 }} $
${m_{tan}}= \frac{e}{2}$
$y - e = \frac{e}{2}(x - 1)$
$y - e = \frac{e}{2}x - \frac{e}{2}$
$y = \frac{e}{2}x - \frac{e}{2} + e$
$y = \frac{e}{2}x + \frac{e}{2}$
Horizontal tangent
$\frac{{{e^{\sqrt x }}}}{{2\sqrt x }} = \frac{0}{1}$
${e^{\sqrt x }} \ne 0$
There are no horizontal tangents.
$f\left( x \right)=~\sqrt{{{x}^{2}}~+4x+5}$ at $x = 0$
$f(x)=(x^2+4x+5)^{ 1/2}$
Point: $\quad f(0) = \sqrt {{0^2} + 4\cdot0 + 5} = \sqrt 5$
$(0,\sqrt 5 )$
$f'(x) = \frac{1}{2}{\left( {{x^2} + 4x + 5} \right)^{ - 1/2}}(2x + 4)$
$f'(x) = \frac{{2x + 4}}{{2\sqrt {{x^2} + 4x + 5} }}$
$f'(x) = \frac{{2(x + 2)}}{{2\sqrt {{x^2} + 4x + 5} }} = \frac{{x + 2}}{{\sqrt {{x^2} + 4x + 5} }}$
Slope: $\quad {m_{tan}}= f'(0) = \frac{{0 + 2}}{{\sqrt {0 + 0 + 5} }} =$
$ {m_{tan}}=\frac{2}{{\sqrt 5 }}$
$y - \sqrt 5 = \frac{2}{{\sqrt 5 }}(x - 0)$
$y - \sqrt 5 = \frac{{2x}}{{\sqrt 5 }}$
$y = \frac{{2x}}{{\sqrt 5 }} + \sqrt 5$
Horizontal tangent
$\frac{{x + 2}}{{\sqrt {{x^2} + 4x + 5} }} = \frac{0}{1}$
$x + 2 = 0$
One horizontal tangent at $\; x = - 2$
Applications
COST FUNCTION: The total cost (in hundreds of dollars) of producing x pairs of sandals per week is: $C\left( x \right) = 6 + \sqrt {3x + 25}$ when $0\; \le x\; \le 30.$
$C(x)=6+{{\left( 3x+25 \right)}^{{1}/{2}\;}}$
$C'(x)=0+\frac{1}{2}{{\left( 3x+25 \right)}^{-1/2}}(3)$
$C'(x)=\frac{3}{2\sqrt{3x+25}}$
$C'(17) = \frac{3}{{2\sqrt {3\cdot17 + 25} }} = .172$
$C'(26) = \frac{3}{{\sqrt {3\cdot26 + 25} }} = .148$
At production level of 17 pairs per week, marginal cost is increasing by \$17.20 per pair.
At production level of 26 pairs per week, marginal cost is increasing by \$14.80 per pair.
PRICE DEMAND EQUATION: The number of large pumpkin spice drinks $(x)$ people are willing to buy per week from a local coffee shop at a price of p (in dollars) is given by:
$$x = 1000 - 60{(p + 25)^{1/2}}$$ when $3.50\; \le p\; \le 6.25.$
$x=1000-60{{\left( p+25 \right)}^{1/2}}$
$\frac{dx}{dp}=0-60(\frac{1}{2}{{\left( p+25 \right)}^{-1/2}})(1)$
$\frac{dx}{dp}=\frac{-30}{\sqrt{p+25}}$
$x = 1000 - 60\sqrt {4.50 + 25} = 674.12$
$\frac{{dx}}{{dp}} = \frac{{ - 30}}{{\sqrt {4.50 + 25} }} = - 5.52$
At \$4.50 the demand is approximately 674 drinks per week. The demand is decreasing by approximately 5 drinks per week.
BIOLOGY: A yeast culture at room temperature (68°F) is placed in a refrigerator set at a constant temperature of 38°F. After t hours, the temperature, T, of the culture is given approximately by
$T = 25{e^{ - 0.62t}} + 38$ $\quad 0\leq t\leq4.$
What is the rate of change of temperature of the culture at the end of 1 hour? At the end of 4 hours?
$T' = 25{e^{ - 0.62t}}( - 0.62)$
$T' = - 15.5{e^{ - 0.62t}}$
$T'(1) = 15.5{e^{ - 0.62(1)}} = - {8.338^ \circ }\;per\;hour$
At the end of one hour, the temperature of the culture is decreasing at a rate of 8.338 degrees per hour.
$T'(4) = - 15.5{e^{( - 0.62(4))}} = - {1.298^ \circ }\;per\;hour$
At the end of five hours, the temperature of the culture is decreasing at at rate of 1.298 degrees per hour.
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}}$
$f'(x) = \frac{1}{2}{\left( {{x^2} - 144} \right)^{ - 1/2}}(2x)$
$f'(x) = \frac{{2x}}{{2\sqrt {{x^2} - 144} }}$
$f'(x) = \frac{x}{{\sqrt {{x^2} - 144} }}$
$g(x) = {e^{2{x^2} - 5x + 4}}$
$g'(x) = {e^{2{x^2} - 5x + 4}}(4x - 5)$
$p(a) = \ln \;({a^4} + 4a)$
$p'(a) = \frac{1}{{{a^4} + 4a}}(4{a^3} + 4)$
$p'(a) = \frac{{4{a^3} + 4}}{{{a^4} + 4a}}$
$y = \frac{1}{{\sqrt[3]{{x - {x^3}}}}}$
$y={{\left( x-{{x}^{3}} \right)}^{-1/3}}$
$y'=-\frac{1}{3}{{\left( x-{{x}^{3}} \right)}^{-4/3}}(1-3{{x}^{2}})$
$y'=\frac{-(1-3{{x}^{2}})}{3{{\left( x-{{x}^{3}} \right)}^{4/3}}}$
$g(x) = 9{(2{x^2} + x - 7)^{ - 3}}$
$g'(x)=-27{{\left( 2{{x}^{2}}+x-7 \right)}^{-4}}(4x+1)$
$g'(x)=\frac{-27(4x+1)}{{{\left( 2{{x}^{2}}+x-7 \right)}^{4}}}$
$y = {(7 - 5\ln x)^3}$
$y'=3\left(7-5\ln x\right)^2\cdot\left(\frac{-5}x\right)$
$y'=\frac{-15\left(7-5\ln x\right)^2}x$
$q(t) = \ln (\ln \;5t)$
$q'\left(t\right)=\frac1{\ln(5t)}\cdot\frac1{5t}\cdot5$
$q'(t) = \frac{1}{{t\ln 5t}}$
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.
$A(21)\approx A(20)+A'(20)$
$A(21)\approx 272.14+2.33 = 274.47$
The total estimated amount of toxic material in the lake at 21 months is $274.47$ ppm.
$p(t) = {t^2}{(5t + 1)^3}$
$f={{t}^{2}}$
$f'=2t$
$g={{\left( 5t+1 \right)}^{3}}$
$g'=3{{\left( 5t+1 \right)}^{2}}(5)$
$g'=15{{\left( 5t+1 \right)}^{2}}$
$p'(t) = 2t{\left( {5t + 1} \right)^3} + {t^2}(15{\left( {5t + 1} \right)^2})$
$p'(t) = 2t{\left( {5t + 1} \right)^3} + 15{t^2}{\left( {5t + 1} \right)^2}$
$p'(t) = {\left( {5t + 1} \right)^2}\left[ {2t(5t + 1) + 15{t^2}} \right]$
$p'(t) = {\left( {5t + 1} \right)^2}\left[ {10{t^2} + 2t + 15{t^2}} \right]$
$p'(t) = {\left( {5t + 1} \right)^2}(25{t^2} + 2t)$
$r(x) = (2{x^2} - 3){(7x + 4)^3}$
$f = 2{x^2} - 3$
$f' = 4x$
$g = {\left( {7x + 4} \right)^3}$
$g' = 3{\left( {7x + 4} \right)^2}(7)$
$g' = 21{\left( {7x + 4} \right)^2}$
$r'(x) = 4x{\left( {7x + 4} \right)^3} + (2{x^2} - 3)\left[ {21{{\left( {7x + 4} \right)}^2}} \right]$
$r'(x) = {\left( {7x + 4} \right)^2}\left[ {4x(7x + 4) + 21(2{x^2} - 3)} \right]$
$r'(x) = {\left( {7x + 4} \right)^2}\left[ {28{x^2} + 16x + 42{x^2} - 63} \right]$
$r'(x) = {\left( {7x + 4} \right)^2}\left[ {70{x^2} + 16x - 63} \right]$
$p(a) = {a^3}\ln(a^5)$
$f = {a^3}$
$f' = 3{a^2}$
$g = \ln ({a^5})$
$g' = \frac{1}{{a^5}} \cdot 5{a^4}$
$g' = \frac5a$
$p'(a) = 3{a^2}\ln {a^5} + {a^3}(\frac5a)$
$p'(a) = 3{a^2}\ln {a^5} + 5a^2$
$p'(a) = a^2(3\ln {a^5} + 5)$
$y =\frac{e^{x^2+x}}{4x-7}$
$f = {e^{{x^2} + x}}$
$f' = {e^{{x^2} + x}}(2x + 1)$
$g = 4x-7$
$g' =4$
$y'=\frac{e^{x^2+x}(2x+1)(4x-7)-4e^{x^2+x}}{\left(4x-7\right)^2}$
$y'=\frac{e^{x^2+x}\left[(2x+1)(4x-7)-4\right]}{\left(4x-7\right)^2}$
$y'=\frac{e^{x^2+x}\left[8x^2-14x+4x-7-4\right]}{\left(4x-7\right)^2}$
$y'=\frac{e^{x^2+x}\left[8x^2-10x-11\right]}{\left(4x-7\right)^2}$
$h(x) = (6x + 5){({x^2} + 4x + 8)^{ - 2}}$
$f = 6x + 5$
$f' = 6$
$g = {\left( {{x^2} + 4x + 8} \right)^{ - 2}}$
$g' = - 2{\left( {{x^2} + 4x + 8} \right)^{ - 3}}(2x + 4)$
$h'(x) = 6{\left( {{x^2} + 4x + 8} \right)^{ - 2}} + (6x + 5)[ - 2{\left( {{x^2} + 4x + 8} \right)^{ - 3}}(2x + 4)]$
$h'(x) = 6{\left( {{x^2} + 4x + 8} \right)^{ - 2}} - 2(6x + 5)(2x + 4){\left( {{x^2} + 4x + 8} \right)^{ - 3}}$
$h'(x) = \frac{6}{{{{\left( {{x^2} + 4x + 8} \right)}^2}}} - \frac{{2(6x + 5)(2x + 4)}}{{{{\left( {{x^2} + 4x + 8} \right)}^3}}}$
$y =\frac{\ln\left(9x+2\right)^2}x$
$f=\ln(9x+2)^2=2\ln\left(9x+2\right)$
$f'=2\cdot\frac1{9x+2}\cdot9=\frac{18}{9x+2}$
$g=x$
$g'=1$
$y'=\frac{\displaystyle\frac{18}{9x+2}\left(x\right)-1\cdot\ln\left(9x+2\right)^2}{x^2}$
$y'=\frac{\displaystyle\frac{18x}{9x+2}-\ln\left(9x+2\right)^2}{x^2}$
Multiply each term by $9x+2.$
$y'=\frac{\displaystyle18x-\left(9x+2\right)\ln\left(9x+2\right)^2}{x^2(9x+2)}$
$q(x) = \frac{e^{2x}}{e^{3x}+1}$
$f = {e^{2x}}$
$f' = {e^{2x}}(2)$
$f' = 2{e^{2x}}$
$g = {e^{3x}} + 1$
$g' = {e^{3x}}(3)$
$g' = 3{e^{3x}}$
$q'(x) = \frac{2e^{2x}(e^{3x}+1)-e^{2x}(3e^{3x})}{(e^{3x}+1)^2}$
$q'(x) = \frac{2e^{5x}+2e^{2x}-3e^{5x}}{\left(e^{3x}+1\right)^2}$
$q'(x) = \frac{-e^{5x}+2e^{2x}}{\left(e^{3x}+1\right)^2}$
$h(x) = \frac{2\sqrt x}{\left(x^2-36\right)^3}$
$f = 2x^{1/2}$
$f' = x^{-1/2}$
$g = (x^2-36)^3$
$g' = 3(x^2-36)^2 \cdot 2x=6x(x^2-36)^2$
$h'(x) = \frac{(x^{-1/2})(x^2-36)^3-6x(x^2-36)^2(2x^{1/2})}{[(x^2-36)^3]^2}$
$h'(x) = \frac{x^{-1/2}(x^2-36)^3-12x^{3/2}(x^2-36)^2}{(x^2-36)^6}$
Factor $(x^2-36)^2$ from each term and reduce.
$h'(x) = \frac{\left(x^2-36\right)^2\left[x^{-1/2}(x^2-36)-12x^{3/2}\right]}{\left(x^2-36\right)^6}$
$h'(x) = \frac{x^{-1/2}(x^2-36)-12x^{3/2}}{(x^2-36)^4}$
Multiply every term by $x^{1/2}$.
$h'(x) = \frac{\left(x^2-36\right)-12x^2}{\sqrt x\left(x^2-36\right)^4}$
$h'(x) = \frac{-11x^2-36}{\sqrt x\left(x^2-36\right)^4}$
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.
$P(5)\approx P(4)+P'(4)$
$P(5)\approx934+263=1197$
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] $