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Mathematics 21 Online
OpenStudy (anonymous):

find the derivative using the chain rule. (problem attached)

OpenStudy (anonymous):

\[f(v) = \left( \frac{1 -2v }{ 1+v } \right)^{3}\]

OpenStudy (anonymous):

ohh it was power of 3!!!!

OpenStudy (anonymous):

grrrrrr

OpenStudy (anonymous):

i have to do it all over again

OpenStudy (anonymous):

haha and you're using the chain rule right?

OpenStudy (anonymous):

yes it is the chain rule but i thought it was to the power of 2

OpenStudy (anonymous):

\[f'(v) = 3*(\frac{ 1-2v }{ 1+v })^2*(\frac{ 1-2v }{ 1+v })'\]

OpenStudy (anonymous):

\[ = 3*(\frac{ 1-2v }{ 1+v })^2*\frac{ -2(1+v)-1(1-2v) }{ (1+v)^2 }\]

OpenStudy (anonymous):

\[ = 3*(\frac{ 1-2v }{ 1+v })^2*\frac{ -2-2v-1+2v }{ (1+v)^2 }\]

OpenStudy (anonymous):

\[ = 3*(\frac{ 1-2v }{ 1+v })^2*\frac{ -3 }{ (1+v)^2 }\]

OpenStudy (anonymous):

\[ = -9*\frac{ (1-2v)^2 }{ (1+v )^4 }\]

OpenStudy (anonymous):

done. lol

OpenStudy (anonymous):

sorry for the misunderstanding at the beginning .. in fact i always see ^3 as ^2 for some reason :| and i already wear glasses!!!

OpenStudy (anonymous):

are you there ? :)

OpenStudy (anonymous):

wow thanks so much for doing all that! and i understand it (: and hahaha dont worry if i didnt have my contacts in i wouldnt be able to read any of this.

OpenStudy (anonymous):

yes.. well the important thing is that you understand :)

OpenStudy (anonymous):

i go to sleep now its 2:49 am in here .. good night :O

OpenStudy (anonymous):

oh wow. its 8:50pm here haha. goodnight! thanks again for your help.

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