dudeperuvian
help me find dy if if y=xlnx then dy=
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vf321
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Product rule.
\[f(x)=x\]\[g(x)=ln(x)\]\[y=f(x)g(x)\]\[dy=(f'(x)g(x)+f(x)g'(x))dx\]
mtaOS
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Yes It seems you will have to use the integration by parts. @vf321 is correctr.
mtaOS
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Did you understand ?
dudeperuvian
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hmm
dudeperuvian
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so it would be (x+lnx)dx
? :3
vf321
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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).
dudeperuvian
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or 1 + lnx dx
mtaOS
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You have to choose which one you integerate and which one you derive.
dudeperuvian
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1lnx+x(...)
vf321
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yes. what is the ...?
dudeperuvian
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(x+lnx)dx
?
vf321
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no... Let's look at it one at a time. What is f(x)?
dudeperuvian
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f(x)= x so f' is equal to one. i get that, but how to derive lnx ;S
dudeperuvian
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thats how i got 1lnx+x(...)
vf321
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That's a definition. d/dx(lnx) = 1/x
vf321
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so now plug into the product rule:
\[dy = (f(x)g'(x) + f'(x)g(x))dx\]
dudeperuvian
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\[ \frac{ lnx+x }{ x}\]
dudeperuvian
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forgot the dx
vf321
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dy = (1+lnx)dx
dudeperuvian
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why?!?!?! -_-
PhoenixFire
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\[f(x) = x\] \[f^{\prime}(x) = 1\]
\[g(x) = \ln(x)\] \[g^{\prime}(x) = {1 \over x}\]
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.