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find f'(x) if f(x) = ln(sinx)
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derivative of sinx divided by sinx
\[\frac{d}{dx}[\ln(f(x))]=\frac{f'(x)}{f(x)}\] via the chain rule
oops
cotx
f(x) = ln(sinx) y' = (1/sinx)(cosx) y' = cosx/sinx y' = cotx
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f'(x)=d(lnsinx)/dx = {d(ln sinx) /dsinx} * {dsinx /dx} ={1/sinx} *{cosx} =cosx/sinx =cotx
got it?
oops? @satellite73
why cot x and not tan x
there was an erroneous tangent floating around, it is gone
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oh look, it came back!!
U can write it interms of tan x too
BUT: f'(x)= 1/tan x
thanks
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