## dudeperuvian Group Title help me find dy if if y=xlnx then dy= 2 years ago 2 years ago

1. vf321 Group Title

Product rule. $f(x)=x$$g(x)=ln(x)$$y=f(x)g(x)$$dy=(f'(x)g(x)+f(x)g'(x))dx$

2. mtaOS Group Title

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

3. mtaOS Group Title

Did you understand ?

4. dudeperuvian Group Title

hmm

5. dudeperuvian Group Title

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

6. vf321 Group Title

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).

7. dudeperuvian Group Title

or 1 + lnx dx

8. mtaOS Group Title

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

9. dudeperuvian Group Title

1lnx+x(...)

10. vf321 Group Title

yes. what is the ...?

11. dudeperuvian Group Title

(x+lnx)dx ?

12. vf321 Group Title

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

13. dudeperuvian Group Title

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

14. dudeperuvian Group Title

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

15. vf321 Group Title

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

16. vf321 Group Title

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

17. dudeperuvian Group Title

$\frac{ lnx+x }{ x}$

18. dudeperuvian Group Title

forgot the dx

19. vf321 Group Title

dy = (1+lnx)dx

20. dudeperuvian Group Title

why?!?!?! -_-

21. PhoenixFire Group Title

$f(x) = x$ $f^{\prime}(x) = 1$ $g(x) = \ln(x)$ $g^{\prime}(x) = {1 \over x}$ 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.