Compound Functions
ExplanationCompound functions are when the output of a one function is the input to another function. To do this one only needs to take the derivative of the outside function and then multiply it by the inside function. The second equation in the chain rule is a way to organize the process word problems and find the connection of one variable to another. For example to find the growth of the volume of bacteria as a function of time (dy/dx) one needs analyze it knowing that volume is a function of radius (dy/dz) and radius is a function of time (dz/dx). Then line them up to match and cross cancel to find dy/dx.
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Chain Rule |
Simple Examples |
Word Problems
The units on the y-axes are millions of people. The x axes increases by 5 years for each tick mark, 0 being 1950.
To find the rate at which the population of Kinshasa is changing, we must find the derivative of
f(x) = 13*(x/13 - 0.103)2 + 0.2
As you will notice, this is a Chain Rule equation. First lets break this up into our inside and outside functions. m(z) = 13(z)2+0.2 z(x) =x/13-0.103 |
Using the Chain rule, we know that f'(x) = m'(z(x))* z'(x). Therefor, using the product rule we can find that:
f'(x) = (13*2) *(x/13 - 0.103)(2-1)* (1*1)x(1-1)/13
f'(x) = 26*(x/13 - 0.103)* 1/13 |
Do you think these numbers accurate?
1965:
1965 -1950 = 15/5 =3
f'(x) = 26*(x/13 - 0.103)* 1/13
f'(3) = 26*((3)/13 - 0.103)* 1/13
f'(3) = 26*(.1277692308)* 1/13
f'(3) = .2555384615
In 1965 the population of Kinshasa was growing at a rate of 0.255,538,461,5 million people per five years.
1990:
1990 -1950 = 40/5 =8
f'(8) = 26*(x/13 - 0.103)* 1/13
f'(8) = 26*((8)/13 - 0.103)* 1/13
f'(8) = 26*(0.5123846154)* 1/13
f'(8) = 1.0247692308
In 1990 the population of Kinshasa was growing at a rate of 1.024,769,230,8 million people per five years.
2040:
2040 -1950 = 90/5 =18
f'(x) = 26*(x/13 - 0.103)* 1/13
f'(18) = 26*((18)/13 - 0.103)* 1/13
f'(18) = 26*( 1.2816153846)* 1/13
f'(18) = 2.5632307692
In 2040 the population of Kinshasa is projected to grow 2.563,230,769,2 million people per five years.
The rate of growth for 1965 seems about correct, because even though it is not on the data line, it seems to be going about parallel at that moment. The growth rate of 1990 seems somewhat inaccurate because even though the line is over-lapping with the data line, it is moving accross it, and not with it. Its slope is much steeper than the data's slope. The growth rate for 2040 seems pretty accurate.
To refresh your memory:
The units on the y-axes are millions of people. The x axes increases by 5 years for each tick mark, 0 being 1950
To find the rate at which Bangkok is growing we must find the derivative of:
q(x) = (1/2)(1/4*x + 1)2 + 1.36
As you will notice, this is a Chain Rule equation. First lets break this up into our inside and outside functions. m(z) = (1/2)(z)2 + 1.36 z(x) =1/4*x + 1 |
Using the Chain rule, we know that q'(x) = m'(z(x))* z'(x). Therefor, using the product rule we can find that:
q'(x) = ((1/2)*2)(1/4*x + 1)(2-1)*(((1/4)*1)*x(1-1) )
q'(x) = (x/4 + 1)*(1/4)
q'(x) = x/16 + 1/4 |
2020:
2020 -1950 = 70/5 = 14
q'(x) = x/16 + 1/4
q'(14) = (14)/16 + 1/4
q'(14) = 0.875 + 1/4
f'(14) = .1.125
In 2020 the population of Bangkok is projected to be growing at a rate of 1.125 million people per year.
2350:
2350 -1950 = 400/5 = 80
q'(x) = x/16 + 1/4
q'(80) = (80)/16 + 1/4
q'(80) = 5 + 1/4
f'(80) = 5.25
In 1990 the population of Kinshasa was growing at a rate of 5.25 million people per year.
The rate of growth for 2020 seems pretty accurate. The projection of 2350 is a little bit extreem. This is assuming a consistent pattern, but since it is very unlikely that Bangkok continuing to grow at such a rate for the next 340 years, the information found above should be taken with a grain of salt.
Cancer
Using the formula for the half life of the Au isotope, assuming that the doctors are still injecting only 10µgrams at a time, here is a formula in which you could find the number of gama rays hitting the cancer as a function of time since injection. We will use that to model the change in rate of the amount of gama rays hitting the cancer over time.
f(t)=.55*(10(1/2)(t/2.7))(1/2)
f'(t) = g'(h (j(t))) * h'(j (t)) * j'(t)
f'(t)=[(.55*(1/2))(10(1/2)(t/2.7))(1/2 - 1)] * [10*ln(1/2)*(1/2)(t/2.7)] * [1/
f'(t)=[(0.275))(10(1/2)(t/2.7))(-1/2)] * [10*ln(1/2)*(1/2)(t/2.7)] * [1/
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f(t)= g(h(j(t))) g=.55*(h(j(t)))(1/2) h=10(1/2)j(t) j=t/2.7
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2 Days
f'(t)=[(0.275))(10(1/2)(t/2.7))(-1/2)] * [10*ln(1/2)*(1/2)(t/2.7)] * [1/ f'(2)=[(0.275))(10(1/2)((2)/2.7))(-1/2)] * [10*ln(1/2)*(1/2)((2)/2.7)] * [1/ f'(2)=[(0.275))(10(1/2)(0.7407407407))(-1/2)] * [10*ln(1/2)*(1/2)(0.7407407407)] * [1/ f'(2)=[(0.275)(10(0.5984320131))(-1/2)] * [10*ln(1/2)*0.5984320131] * [0.3703703704]
f'(2)=[(0.275)(5.984320131)(-1/2)] * [-6.931471806*0.5984320131] * [0.3703703704]
f'(2)=[(0.275)(5.984320131)(-1/2)] * [-6.931471806*0.5984320131] * [0.3703703704]
f'(2)=[(0.275)(0.4087827783)] * [-4.1480146266] * [0.3703703704]
f'(2)=0.112415264 * -4.1480146266 * 0.3703703704
f'(2)=-0.1727037627.
2 days after injection, the quantity of gama rays is going down by -0.1727037627 per day.