# chain rule

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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...

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...

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#4

(Original post by

use two substitutions to find

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

thanks!

**sarah12345**)use two substitutions to find

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

thanks!

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)]

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(Original post by

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)]

**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)]

one thing i dont understand is why do you have to mave 2 substitutions why not just say that u=srt(1-x^2)??????

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(Original post by

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)??????

**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)??????

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#7

**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)??????

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(Original post by

Because the question asked me to use 2 substitutions, so I did.

**Nima**)Because the question asked me to use 2 substitutions, so I did.

thanks alot for all the help

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