# chain rule

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#1
use two substitutions to find

d/dx (e^rt(1-x^2))

thanks!
0
15 years ago
#2
u = 1-(x^2)
v = u^(1/2)
0
15 years ago
#3
chain rule...

dy/dx = (dy/du)(du/dx)

however, as we are doing two substitutions, we'll need....

dy/dx = (dy/du)(du/dx)
dy/du = (dy/dv)(dv/du)

dy/dx = (dy/dv)(dv/du)(du/dx)

y = (e^rt(1-x^2))
u = 1-(x^2)
v = u^(0.5)

what we need to differentiate...

y = e^v
v = u^(0.5)
u = 1-(x^2)

dy/dv = (e^v)
dv/du = (0.5)u^-0.5
du/dx = -2x

dy/dx = (dy/dv)(dv/du)(du/dx)
dy/dx = [e^{(1-x^2)^0.5)}][0.5{(1-x^2)^-0.5}][-2x]

be as well getting someone else to check...
0
15 years ago
#4
(Original post by sarah12345)
use two substitutions to find

d/dx (e^rt(1-x^2))

thanks!
Q.) Find d/dx [e^rt(1 - x^2)]
Let the function we want to differentiate be y = [e^Sqrt(1 - x^2)]

Sub 1.) Let u = (1 - x^2) ---> du/dx = -2x
Sub 2.) Let v = Sqrt(u) = u^1/2 ---> dv/du = 1/2.u^(-1/2) = 1/[2Sqrt(u)]
Hence: y = e^v ---> dy/dv = e^v

Hence: d/dx [e^Sqrt(1 - x^2)] = dy/dx = dy/dv * dv/du * du/dx = e^v * 1/[2Sqrt(u)] * (-2x) = -{2xe^[Sqrt(1 - x^2)]}/[2Sqrt(1 - x^2)]

---> d/dx [e^rt(1 - x^2)] = -{xe^[Sqrt(1 - x^2)]}/[Sqrt(1 - x^2)]
0
#5
(Original post by Nima)
Q.) Find d/dx [e^rt(1 - x^2)]
Let the function we want to differentiate be y = [e^Sqrt(1 - x^2)]

Sub 1.) Let u = (1 - x^2) ---> du/dx = -2x
Sub 2.) Let v = Sqrt(u) = u^1/2 ---> dv/du = 1/2.u^(-1/2) = 1/[2Sqrt(u)]
Hence: y = e^v ---> dy/dv = e^v

Hence: d/dx [e^Sqrt(1 - x^2)] = dy/dx = dy/dv * dv/du * du/dx = e^v * 1/[2Sqrt(u)] * (-2x) = -{2xe^[Sqrt(1 - x^2)]}/[2Sqrt(1 - x^2)]

---> d/dx [e^rt(1 - x^2)] = -{xe^[Sqrt(1 - x^2)]}/[Sqrt(1 - x^2)]
thanks alot!!
one thing i dont understand is why do you have to mave 2 substitutions why not just say that u=srt(1-x^2)??????
0
#6
(Original post by sarah12345)
thanks alot!!
one thing i dont understand is why do you have to mave 2 substitutions why not just say that u=srt(1-x^2)??????
anyone?
0
15 years ago
#7
(Original post by sarah12345)
thanks alot!!
one thing i dont understand is why do you have to mave 2 substitutions why not just say that u=srt(1-x^2)??????
That works if you can differentiate sqrt(1 - x^2) in your head - a useful skill!
0
#8
(Original post by Nima)
Because the question asked me to use 2 substitutions, so I did.
ok thanks its just i was wonderin if there were certain times when u had to do two substiituitions for it to work.

thanks alot for all the help
0
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