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help me find dy if if y=xlnx then dy=

Mathematics
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Product rule. \[f(x)=x\]\[g(x)=ln(x)\]\[y=f(x)g(x)\]\[dy=(f'(x)g(x)+f(x)g'(x))dx\]
Yes It seems you will have to use the integration by parts. @vf321 is correctr.
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Other answers:

hmm
so it would be (x+lnx)dx ? :3
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).
or 1 + lnx dx
You have to choose which one you integerate and which one you derive.
1lnx+x(...)
yes. what is the ...?
(x+lnx)dx ?
no... Let's look at it one at a time. What is f(x)?
f(x)= x so f' is equal to one. i get that, but how to derive lnx ;S
thats how i got 1lnx+x(...)
That's a definition. d/dx(lnx) = 1/x
so now plug into the product rule: \[dy = (f(x)g'(x) + f'(x)g(x))dx\]
\[ \frac{ lnx+x }{ x}\]
forgot the dx
dy = (1+lnx)dx
why?!?!?! -_-
\[f(x) = x\] \[f^{\prime}(x) = 1\] \[g(x) = \ln(x)\] \[g^{\prime}(x) = {1 \over x}\] 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.

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