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derivate (x)/(((x^2)+4)^2)

\(\begin{array}{lcl} y = \frac{x}{(4+x^2)^2} &=& \frac{x}{(4+x^2)\cdot (4+x^2)} \end{array} \)

Use the general quotient rule:

\(\begin{array}{lcll} y & = & \frac{f(x)}{ g(x)\cdot h(x) }\\ y' &=& y\cdot \left[ \frac{ f'(x) } {f(x)} - \frac{ g'(x) }{g(x)} -\frac{ h'(x) }{h(x)} \right] \\ y' &=& \frac{f(x)}{ g(x)\cdot h(x) }\cdot \left[ \frac{ f'(x) } {f(x)} - \frac{ g'(x) }{g(x)} -\frac{ h'(x) }{h(x)} \right] \\ y' &=& \frac{ f'(x) } { g(x)\cdot h(x)} - \frac{ f(x)\cdot g'(x) }{g^2(x)\cdot h(x)} -\frac{ f(x)\cdot h'(x) }{g(x)\cdot h^2(x)} \\ y' &=& \frac{ f'(x)\cdot g(x)\cdot h(x) - f(x)\cdot g'(x)\cdot h(x) - f(x)\cdot h'(x) \cdot g(x) } {g^2(x)\cdot h^2(x)} \\\\ \end{array}\\ \begin{array}{lcll} f(x) &=& x & f'(x) = 1 \\ g(x) &=& 4+x^2 & g'(x) = 2x \\ h(x) &=& 4+x^2 & h'(x) = 2x \\\\ \end{array}\\ \begin{array}{lcll} y' &=& \frac{ 1\cdot (4+x^2)\cdot (4+x^2) - x\cdot 2x\cdot (4+x^2) - x\cdot 2x \cdot (4+x^2) } {(4+x^2)^2\cdot (4+x^2)^2} \\ y' &=& \frac{ (4+x^2)^2 - 2\cdot [~ x\cdot 2x\cdot (4+x^2)~] } {(4+x^2)^4} \\ y' &=& \frac{ (4+x^2)^2 - 4\cdot x^2\cdot (4+x^2) } {(4+x^2)^4} \\ y' &=& \frac{ (4+x^2)^2 }{(4+x^2)^4} - \frac{ 4\cdot x^2\cdot (4+x^2) } {(4+x^2)^4} \\ y' &=& \frac{ 1 }{(4+x^2)^2} - \frac{ 4\cdot x^2 } {(4+x^2)^3} \\ \end{array}\)

Explain how different quadratic functions can have the same zeros.

\(\begin{array}{lcl} \text{Quadratic functions with same zeroes are: } a(x-x_1)(x-x_2) = 0 \\ \quad \text{ or } \quad ax^2 + [ -a\cdot (x_1+x_2)] x + a\cdot x_1\cdot x_2 = 0\\ \text{the zeroes are } x_1 \text{ and } x_2 \text{ and } a\ne 0\\\\ \hline \\ \text{Example 1: }\quad a=1\qquad y = x^2+bx+c = 0 \\ b=-(x_1+x_2) \qquad c=x_1\cdot x_2 \qquad \text{the zeroes are } x_1 \text{ and } x_2\\\\ \text{Set } x = x_1: \\ y =(x_1)^2 + [-(x_1+x_2)]\cdot x_1 + x_1\cdot x_2 \\ y =(x_1)^2 - (x_1)^2 - x_2\cdot x_1 + x_1\cdot x_2 = 0\\\\ \text{Set } x = x_2: \\ y =(x_2)^2 + [-(x_1+x_2)]\cdot x_2 + x_1\cdot x_2 \\ y =(x_2)^2 - x_1\cdot x_2 - (x_2)^2 + x_1\cdot x_2 = 0 \\ \hline \\ \text{Example 2: }\quad a=2\qquad y = 2x^2+bx+c = 0 \\ b=-2\cdot (x_1+x_2) \qquad c=2\cdot x_1\cdot x_2 \qquad \text{the zeroes are } x_1 \text{ and } x_2\\\\ \text{Set } x = x_1: \\ y =2(x_1)^2 + [-2(x_1+x_2)]\cdot x_1 + 2x_1\cdot x_2 \\ y =2(x_1)^2 - 2(x_1)^2 - 2x_2\cdot x_1 + 2x_1\cdot x_2 = 0\\\\ \text{Set } x = x_2: \\ y =2(x_2)^2 + [-2(x_1+x_2)]\cdot x_2 + 2x_1\cdot x_2 \\ y =2(x_2)^2 - 2x_1\cdot x_2 - 2(x_2)^2 + 2x_1\cdot x_2 = 0 \end{array}\)

\(\begin{array}{lcl} \text{Quadratic functions with same zeroes are: }\\\\ \end{array}\\ \begin{array}{rcrcl} && 2x^2+2.2x+0.2&=&0 \qquad \text{the zeroes are } -1 \text{ and } -0.1 \\ 2 &\cdot & (2x^2+2.2x+0.2)=4x^2+4.4x+0.4&=&0 \qquad \text{the zeroes are } -1 \text{ and } -0.1 \\ -2&\cdot & (2x^2+2.2x+0.2)=-4x^2-4.4x-0.4&=&0 \qquad \text{the zeroes are } -1 \text{ and } -0.1 \\ 1.5&\cdot & (2x^2+2.2x+0.2)=3x^2+3.3x+0.3&=&0 \qquad \text{the zeroes are } -1 \text{ and } -0.1 \\ \cdots \end{array}\)

https://www.schlaukopf.de/data/images/45631.png ich kann sie nicht lösen 

\(\begin{array}{rcl} \frac{7}{18} + \frac{8}{27} + \frac{1}{6} &=& \frac{7}{3\cdot 6} + \frac{8}{3\cdot 9} + \frac{1}{3\cdot 2}\\ &=& \frac{1}{3} \cdot \left[ \frac{7}{6} + \frac{8}{9} + \frac{1}{2} \right]\\ &=& \frac{1}{3} \cdot \left[ \frac{7}{6}\cdot \frac{3}{3} + \frac{8}{9}\cdot \frac{2}{2} + \frac{1}{2}\cdot \frac{9}{9} \right]\\ &=& \frac{1}{3} \cdot \left[ \frac{21}{18} + \frac{16}{18} + \frac{9}{18} \right]\\ &=& \frac{1}{3} \cdot \frac{46}{18}\\ &=& \frac{1}{3} \cdot \frac{23}{9}\\ &=& \frac{23}{27}\\ \end{array}\)

Find a formula for an for the arithmetic sequence a3 = -10, a7 = -18

\(\begin{array}{rcll} \boxed{~ a_n = a_1 + (n-1)d ~ }\\ a_3 &=& a_1 + (3-1)d = -10\\ a_7 &=& a_1 + (7-1)d = -18\\ a_3-a_7 = -10-(-18) &=& a_1 + (3-1)d - [~a_1 + (7-1)d~] \\ -10+18 &=& a_1 + 2d - a_1 - 6d \\ 8 &=& 2d-6d \\ 8 &=& -4d\\ 8 &=& -4d \qquad | \qquad : -4\\ -2 &=& d \\ \mathbf{d} &\mathbf{=}& \mathbf{-2} \\\\ a_3 &=& a_1 + 2d \\ a_1 &=& a_3-2d \qquad a_3=-10 \qquad d = -2\\ a_1 &=& -10-2\cdot (-2) \\ a_1 &=& -10+4 \\ \mathbf{a_1} &\mathbf{=}& \mathbf{-6} \end{array}\)

\(\text{Formula}\\ \boxed{~ \begin{array}{lcll} a_n &=& -6 + (n-1)\cdot(-2) \\ a_n &=& -6 -2n+2\\ a_n &=& -4-2n \end{array} ~}\)

sqrt(1-x^2)-x/2 What are the 0 points of the derivative function?

\(\begin{array}{rcll} y = \sqrt{1-x^2}- \frac{x}{2} &=& (1-x^2)^{\frac12} -\frac{x}{2}\\\\ y' &=& \frac12 \cdot (1-x^2)^{\frac12 -1} \cdot (-2x) - \frac12 \\ y' &=& \frac12 \cdot (1-x^2)^{-\frac12} \cdot (-2x) - \frac12 \\ y' &=& (1-x^2)^{-\frac12} \cdot (-x) - \frac12 \\ y' &=& -\frac{x}{ \sqrt{1-x^2} }- \frac12 \\ y'=0\\ -\frac{x}{ \sqrt{1-x^2} }- \frac12 &=& 0 \\ -\frac{x}{ \sqrt{1-x^2} } &=& \frac12 \\ -2x &=& \sqrt{1-x^2} \qquad | \qquad ()^2\\ (-2x)^2 &=& (\sqrt{1-x^2})^2 \\ 4x^2 &=& 1-x^2\\ 5x^2 &=& 1\\ x^2 &=& \frac15 \qquad | \qquad \sqrt{} \\ x_{1,2} &=& \pm \sqrt{\frac15} \\\\ x_1 &=& \sqrt{\frac15}\\ x_1 &=& \frac{1}{\sqrt{5} } \qquad \text{ no solution!}\\\\ x_2 &=& -\sqrt{\frac15} \\ x_2 &=& -\frac{1}{\sqrt{5} } \end{array}\)

\(\begin{array}{rcll} \text{The solution is } \quad x = -\frac{1}{\sqrt{5} } \end{array}\)

If a cat with a mass of 4.4 kilograms can run at a velocity of 12.5m/s how much kinetic energy does the cat produce

\(\begin{array}{rcll} E_k &=& \frac12 m v^2 \qquad m = 4.4\ \mathrm{kg} \qquad v = 12.5\ \frac{\mathrm{m}}{\mathrm{s}} \\\\ E_k &=& \frac12 \cdot 4.4\ \mathrm{kg} \cdot \left(12.5\ \frac{\mathrm{m}}{\mathrm{s}} \right)^2 \\ E_k &=& 2.2 \cdot 156.25\ \frac{\mathrm{kg}\cdot \mathrm{m}^2}{\mathrm{s}^2} \\ E_k &=&343.75\ \frac{\mathrm{kg}\cdot \mathrm{m}^2}{\mathrm{s}^2} \\ E_k &=&343.75\ \mathrm{J} \end{array}\)

Solve for K Wt=Wi*e^-kt

I assume: \(W(t) = W(i) \cdot e^{-kt}\)

\(\begin{array}{rcll} W(t) &=& W(i) \cdot e^{-kt} \qquad &| \qquad : W(i) \\ \frac{W(t)}{ W(i) } &=& e^{-kt} \qquad &| \qquad \ln{} \\ \ln{ (\frac{W(t)}{ W(i) }) } &=& \ln{ (e^{-kt}) } \\ \ln{ (W(t)) }-\ln{ (W(i)) } &=& -kt\cdot \ln{ (e ) } \qquad &| \qquad \ln{(e)} = 1 \\ \ln{ (W(t)) }-\ln{ (W(i)) } &=& -kt \qquad &| \qquad \cdot(-1) \\ -\ln{ (W(t)) }+\ln{ (W(i)) } &=& kt \\ kt &=& -\ln{ (W(t)) }+\ln{ (W(i)) }\\ kt &=& \ln{ (W(i)) } - \ln{ (W(t)) } \qquad &| \qquad : t \\ \mathbf{k} &\mathbf{=}& \mathbf{\frac{\ln{ (W(i)) } - \ln{ (W(t)) }}{t} } \\ \end{array}\)

Yellowstone National Park is a popular field trip destination. This year the senior class at High School A and the senior class at High School B both planned trips there. The senior class at High School A rented and filled 9 vans and 4 buses with 256 students. High School B rented and filled 8 vans and 8 buses with 392 students. Every van had the same number of students in it as did the buses. Find the number of students in each van and in each bus.

v = van

b = bus

\(\begin{array}{lcl} A && B \\ \hline 9\text{ vans} && 8\text{ vans} \\ 4\text{ buses} && 8\text{ buses} \\ \hline =256 \text{ students} && =392 \text{ students} \end{array}\)

\(\begin{array}{lrclclrcl} (1)& 9v+4b &=& 256 \quad |\quad:4 && (2) & 8v+8b &=& 392 \quad |\quad : 8 \\ &2.25v+b&=&64 && & v+b &=& 49 \\ \hline \\ (1)-(2) & 2.25v+b - (v+b) &=& 64-49\\ & 2.25v+b -v-b &=& 15\\ & 1.25v &=& 15 \quad |\quad : 1.25 \\ & v &=& \frac{15}{ 1.25} \\ & \mathbf{v} &\mathbf{=}& \mathbf{12} \\ \hline \\ (2) & v+b &=& 49 \\ & 12+b &=& 49 \quad |\quad -12 \\ & b &=& 49-12\\ & \mathbf{b} &\mathbf{=}& \mathbf{37} \end{array}\)

Check:

9*12 + 4*37 = 256   Okay

8*12 + 8*37 = 392   Okay

A scientist counts 25 bacteria present in a culture and finds that the number of bacteria triples each hour. The function y=25 times 3x models the number of bacteria after x hours. Estimate when there will be about 1170 bacteria in the culture.

\(\begin{array}{rcl} y &=&25 \cdot 3^x \\ 1170 &=&25 \cdot 3^x \\ 25\cdot 3^x &=& 1170 \qquad & |\qquad : 25\\ 3^x &=& \frac{1170}{ 25 } \\ 3^x &=& 46.8 \qquad & |\qquad \log_{10}{}\\ \log_{10}{(3^x)} &=& \log_{10}{(46.8)} \\ x\cdot \log_{10}{(3)} &=& \log_{10}{(46.8)} \qquad & |\qquad : \log_{10}{(3)} \\ x &=& \frac{ \log_{10}{(46.8)} } { \log_{10}{(3)} }\\ x &=& \frac{1.67024585307} { 0.47712125472 }\\ \mathbf{x} &\mathbf{=}& \mathbf{3.50067375233} \end{array}\)

about 3.5 hours