Question

Differentiate:

M(t)= [p/1-(1-p)e^t]^k

Answer 1

\[M(t)=[\frac p{1-(1-p)e^t}]^k\]is the formula for M ?

Answer 2

yep thats the formula i want to differentiate

Answer 3

and p and k are constants, right?

Answer 4

yes

Answer 5

Chain Rule\[(f\circ g)'(x)=f'(g(x))\cdot g'(x)\]In your case we can define the functions as so:\[f(x)=x^k\]\[g(x)=\frac p{1-(1-p)e^t}\]f can be done with the product rule, and g can be done with the quotient rule.

Find each derivative and plug it into the product rule.

Answer 6

ok thank you very much..so u positive this is the right method to go about doing this?

Answer 7

Yes
actually you can do the second part as\[g(x)=p[1-(1-p)e^t]^{-1}\]if you would rather use the chain rule again instead of the product rule, but you will get the same answer either way

Answer 8

I meant instead of the quotient rule*

Answer 9

ok il give it a try now

Answer 10

Cool, let me know if you want me to post how I did it...

Answer 11

yes please

Answer 12

il be grateful

Answer 13

\[M(t)=[\frac p{1-(1-p)e^t}]^k\]\[M'(t)=k[\frac p{1-(1-p)e^t}]^{k-1}\cdot(-p)[1-(1-p)e^t]^{-2}\cdot(p-1)e^t\]Now simplify...

Answer 14

so u used the chain rule?

Answer 15

yeah, twice

Answer 16

\[f(t)=t^k\]\[g(t)=pt^{-1}\]\[h(t)=1-(1-p)e^t\]take each derivative and plug into the chain rule\[(f\circ g\circ h)'(t)=(f'\circ g\circ h)(t)\cdot(g'\circ h)(t)\cdot h'(t)\]sorry for changing t to x earlier, I hope that didn't confuse you

Answer 17

so theres thhees functions

Answer 18

three*

Answer 19

there doesn't have to be, I just renamed the parts to make it easier

you can also just say\[f(t)=t^k\]and\[g(t)=\frac p{1-(1-p)e^t}\]that way you only have two functions, but you will have to use the product or quotient rule on the second one

Answer 20

Thank You very much..it seems that ur a pro at differentiation

Answer 21

Differentiation is easy once you get used to it.
Integration, on the other hand, remains tricky for much longer ;)

Answer 22

sorry...how do u differentiate 1-(1-p)e^t

Answer 23

the first one is a constant, so that becomes zero
we can treat (1-p) as a constant coefficient, and d/dx(e^x)=e^x, so we get\[f(t)=1-(1-p)e^t\]\[f'(t)=-(1-p)e^t\]

Answer 24

so the minus sign stays

Answer 25

Yes, I think above I changed it like so:\[-(1-p)e^t=(p-1)e^t\]sorry if that threw you
if you don't trust that analysis, you can distribute first\[1-
(1-p)e^t=1-e^t+pe^t\]now take the derivative:\[-e^t+pe^t=(p-1)e^t\]so it works out the same either way, as it should

Answer 26

ok thanks

Answer 27

:)

Answer 28

welcome :D

Answer 29

after im done can u quickly check my answer for me please ?

Answer 30

sure, but so can wolfram

Answer 31

?

Answer 32

There are a lot of different ways to simplify this answer, but mine came out looking like the first one under 'alternate forms'
http://www.wolframalpha.com/input/?i=d%2Fdt%5Bp%2F%281-%281-p%29e%5Et%29%5D%5Ek

Answer 33

If you are not familiar with wolfram, it is a good way to check answers.
Don't take it as gospel though, I've caught making mistakes many times

Answer 34

how would u simplify: [1-(1-p)e^t -p(p-1)e^t] / (1-(1-p)e^t)

Answer 35

sorry

Answer 36

thats a mistake

Answer 37

[1-(1-p)e^t -p(p-1)e^t] / (1-(1-p)e^t)^2

Answer 38

how would u simplify the above

Answer 39

I would change the numerator to\[1-(1-p)e^t +p(1-p)e^t\]and factor out (1-p)e^t:\[(1-1+p)(1-p)e^t=p(1-p)e^t\]and leave the denominator the same

Answer 40

oh I think I messed up there XD
on sec...

Answer 41

cant u just get rid of: 1-(1-p)e^t from numerator and denominator?

Answer 42

[1-(1-p)e^t -p(p-1)e^t]= [1-(1-p)e^t +p(1-p)e^t]=1-(1-p)(1-p)e^t
but if that is your answer I think you are missing something...

the way you have it written you could make it
1-[p(p-1)e^t] / (1-(1-p)e^t)^2
but it's hard to match up our answers because there are a lot of symbols
I'll type my simplification...

Answer 43

\[M'(t)=k[\frac p{1-(1-p)e^t}]^{k-1}\cdot(-p)[1-(1-p)e^t]^{-2}\cdot(p-1)e^t\]\[=\frac{pk(1-p)e^t[\frac p{1-(1-p)e^t}]^{k-1}}{[1-(1-p)e^t]^2}\]which is the same as wolfram's answer (the first one under 'Alternate forms')

Answer 44

wolfram has a few +/- bits different than me, but you can reconcile that without too much difficulty through algebra

they are the same

Answer 45

...as far as how wolfram simplified to the answer at the top of the page, I don't feel like figuring that out. Seems excessive.

Answer 46

no thats not my final answer

Answer 47

i just need the simplification of what i gave above

Answer 48

well without context it still looks wrong to me, but it is hard to tell

I have to figure out what exactly you isolated in that part of the answer, and that's kinda hard to do from here...

Answer 49

is that part you are asking about just what you got from the quotient rule?

Answer 50

simplify: [1-(1-p)e^t + p(1-p)e^t]/ 1-(1-p)e^t)^2

Answer 51

yh

Answer 52

yes from quotient rule

Answer 53

I get, by factoring out (1-p)e^t\[\frac{1-(1-p)e^t + p(1-p)e^t}{ (1-(1-p)e^t)}^2=\frac{1+(1-p)(1-p)e^t}{1-(1-p)e^t}\]but I'm not so sure you did the quotient rule right, let me check

Answer 54

oops

Answer 55

sorry the denomniator has a squared not the numerator

Answer 56

so the denomaitor is: (1-(1-p)e^t)^2

Answer 57

\[\frac p{1-(1-p)e^t}\]\[\frac{0-p[-(1-p)]e^t}{[1-(1-p)e^t]^2}=\frac{p(1-p)e^t}{[1-(1-p)e^t]^2}\]so I think you messed up on the derivative a bit

Answer 58

...and that, times k[p/(1-(1-p)e^t)]^(k-1) gives my answer

Answer 59

u=p, du/dp=1....v=1-(1-p)e^t, dv/dp=-(1-p)e^t..is that how i start?

Answer 60

p is a constant though...

Answer 61

dp/dt=0

Answer 62

we are not taking the derivative w/respect to p

Answer 63

u=p
du/dt=0

v is correct though

Answer 64

o silly me...sorry so it becomes zero lol?

Answer 65

because p is constant I prefer to rewrite it as\[p[1-(1-p)e^t]^{-1}\]and use the chain rule instead. less messy in my opinion, but to each their own

Answer 66

yes to your last question

Answer 67

i dont know how to use chain rule with the above i get confused

Answer 68

and v=-(1-p)e^t

Answer 69

^that is dv/dt you mean
here it is with the chain rule:\[p[1-(1-p)e^t]^{-1}=(-1)\cdot p[1-(1-p)e^t]^{-2}\cdot[-(1-p)e^t]\]\[=\frac{p(1-p)e^t}{[1-(1-p)e^t]^2}\]again, same answer

Answer 70

thank you very much...sorry for my clumsiness

Answer 71

Not at all, it's just a matter of practice ;)
again, welcome!

Answer 72

my final answer:

k[p(1-p)e^t/[1-(1-p)e^t]^2] [p/1-(1-p)e^t]^k-1

Answer 73

does this look promising?

Answer 74

yes, I think that's right :D

Answer 75

at last :D

Answer 76

\[\frac{kp(1-p)e^t}{[1-(1-p)e^t]^2} \cdot[{1-(1-p)e^t}]^{k-1 }\]yeah that simplifies to my answer

Answer 77

whoops dropped the p from the numerator, but you are right, don't worry

Answer 78

LOL

Answer 79

\[\frac{kp(1-p)e^t}{[1-(1-p)e^t]^2} \cdot[\frac p{1-(1-p)e^t}]^{k-1 }\]there we go, that's just a stone throw away from mine
nicely done, congrads!

Answer 80

Thank You...appreciate it

Answer 81

sorry to have bothered u too much.

Answer 82

Not at all, that's why I'm here

Answer 83

:D

Answer 84

what methods would you use if we were to take the econd derivative of the answer we achieved above? so M''(t)?

Answer 85

It's not going to be pretty...

In that case it would be nice to get to wolframs factored form, then you could use the product rule.
however with what we have, what I would probably do is write\[\frac{kp(1-p)e^t}{[1-(1-p)e^t]^2} \cdot[\frac p{1-(1-p)e^t}]^{k-1 }\]\[u=\frac{kp(1-p)e^t}{[1-(1-p)e^t]^2}\]\[v=[\frac p{1-(1-p)e^t}]^{k-1 }\]\[M''(t)=u'v+v'u\]no note that v' will be the same as the derivative we just found, but with k-1 instead of k. This way we can save a lot of work.
That means you only need to find u', which can be done with the quotient rule.
Note that you will need the chain rule on the denominator though, so yes, ugly answer expected.

Answer 86

actually it's not as bad as I thought after simplification
http://www.wolframalpha.com/input/?i=d%5E2%2Fdt%5E2%5Bp%2F%281-%281-p%29e%5Et%29%5D%5Ek
I gotta eat, I'll check back later.
Good luck

Answer 87

also in finding u' you can take the kp(1-p) out (since it is constant) and just find the derivative of\[\frac{e^t}{[1-(1-p)e^t]^2}\]which should be a little less messy

just don't forget to add back in kp(1-p) at the end