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vf321Best ResponseYou've already chosen the best response.2
Product rule. \[f(x)=x\]\[g(x)=ln(x)\]\[y=f(x)g(x)\]\[dy=(f'(x)g(x)+f(x)g'(x))dx\]
 one year ago

mtaOSBest ResponseYou've already chosen the best response.0
Yes It seems you will have to use the integration by parts. @vf321 is correctr.
 one year ago

dudeperuvianBest ResponseYou've already chosen the best response.0
so it would be (x+lnx)dx ? :3
 one year ago

vf321Best ResponseYou've already chosen the best response.2
NO! I made up some functions, f(x) and g(x). Can you tell me what they are? (Hint: Look up). Then find f'(x) and g'(x).
 one year ago

mtaOSBest ResponseYou've already chosen the best response.0
You have to choose which one you integerate and which one you derive.
 one year ago

vf321Best ResponseYou've already chosen the best response.2
no... Let's look at it one at a time. What is f(x)?
 one year ago

dudeperuvianBest ResponseYou've already chosen the best response.0
f(x)= x so f' is equal to one. i get that, but how to derive lnx ;S
 one year ago

dudeperuvianBest ResponseYou've already chosen the best response.0
thats how i got 1lnx+x(...)
 one year ago

vf321Best ResponseYou've already chosen the best response.2
That's a definition. d/dx(lnx) = 1/x
 one year ago

vf321Best ResponseYou've already chosen the best response.2
so now plug into the product rule: \[dy = (f(x)g'(x) + f'(x)g(x))dx\]
 one year ago

dudeperuvianBest ResponseYou've already chosen the best response.0
\[ \frac{ lnx+x }{ x}\]
 one year ago

PhoenixFireBest ResponseYou've already chosen the best response.1
\[f(x) = x\] \[f^{\prime}(x) = 1\] \[g(x) = \ln(x)\] \[g^{\prime}(x) = {1 \over x}\] Product rule: \[{{\delta fg}\over{\delta y}} = f^{\prime}g + fg^{\prime}\] Now substitute the equations calculated above. \[{\delta y \over \delta x} = 1\ln(x) + x{1 \over x}\] \[{\delta y \over \delta x} = \ln(x) + 1\] the x * 1/x is the same as x/x, which is the same as 1.
 one year ago
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