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Mathematics 15 Online
OpenStudy (anonymous):

derivatives

OpenStudy (anonymous):

@gorv help!!

OpenStudy (anonymous):

@gorv

OpenStudy (gorv):

what is derivative of ln x??

OpenStudy (gorv):

@mondona

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

1

OpenStudy (mrnood):

The Chain Rule dy/dx = dy/du *du/dx let u= (1-2x) then use the rule above....

OpenStudy (anonymous):

i got 1/(1-2x)^2 thank you(:

OpenStudy (anonymous):

thats right, right?

OpenStudy (johnweldon1993):

Not sure where you got your exponent from...

OpenStudy (mrnood):

doesn't look right to me - haven't done th esum - but ther is prolly a 2* something in there from du/dx

OpenStudy (johnweldon1993):

^Indeed there will be a "times" 2 because you need to take the derivative of the inside function as well

OpenStudy (dumbcow):

\[\frac{d}{du} \ln u = \frac{1}{u}\] \[\frac{d}{dx} (1-2x) = -2\]

OpenStudy (anonymous):

ohh so it would be 1-2x sorry!

OpenStudy (anonymous):

1/1-2*

OpenStudy (anonymous):

1/1-2x* lol my bad

OpenStudy (mrnood):

slow down - do the process methodically

OpenStudy (mrnood):

@dumbcow has given you nearly all you need

OpenStudy (dumbcow):

\[\frac{dy}{dx} = \frac{1}{u}* (-2) = \frac{-2}{1-2x}\]

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