Question

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

Answer 1

Product rule.
\[f(x)=x\]\[g(x)=ln(x)\]\[y=f(x)g(x)\]\[dy=(f'(x)g(x)+f(x)g'(x))dx\]

Answer 2

Yes It seems you will have to use the integration by parts. @vf321 is correctr.

Answer 3

Did you understand ?

Answer 4

hmm

Answer 5

so it would be (x+lnx)dx
? :3

Answer 6

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).

Answer 7

or 1 + lnx dx

Answer 8

You have to choose which one you integerate and which one you derive.

Answer 9

1lnx+x(...)

Answer 10

yes. what is the ...?

Answer 11

(x+lnx)dx
?

Answer 12

no... Let's look at it one at a time. What is f(x)?

Answer 13

f(x)= x so f' is equal to one. i get that, but how to derive lnx ;S

Answer 14

thats how i got 1lnx+x(...)

Answer 15

That's a definition. d/dx(lnx) = 1/x

Answer 16

so now plug into the product rule:
\[dy = (f(x)g'(x) + f'(x)g(x))dx\]

Answer 17

\[ \frac{ lnx+x }{ x}\]

Answer 18

forgot the dx

Answer 19

dy = (1+lnx)dx

Answer 20

why?!?!?! -_-

Answer 21

\[f(x) = x\] \[f^{\prime}(x) = 1\]
\[g(x) = \ln(x)\] \[g^{\prime}(x) = {1 \over x}\]

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.