monroe17 2 years ago Find dy/dx step by step? please? :)

1. monroe17

$y=ln(\frac{ x^4e^{3x} }{ \sqrt{2x^5-1} })$

2. dpaInc

yuck!!!

3. monroe17

I know ;(

4. dpaInc

i think it would be better to do logarithmic differentiation...

5. dpaInc

$$\large y=ln(\frac{x^4e^{3x}}{(2x-5)^{\frac{1}{2}}}) = 4lnx + 3x - \frac{1}{2}(2x^5-1)$$ simplifying y that way should be much simpler...

6. dpaInc

i guess that's not so hard afterall....

7. monroe17

wait wait.. how'd you get 4ln(x)+3x-1/2(2x^5-1)

8. dpaInc

use the logaritm properties...

9. monroe17

I need to look those up ;/ and review them.

10. dpaInc

$$\large ln(AB)=lnA+lnB$$; $$\large ln(A/B)=lnA-lnB$$; $$\large ln(A^B)=BlnA$$; hmm... i think that's all of 'em...

11. monroe17

Awe, well thank you. Okay so now back to what you got.. Let me look over that real quick

12. dpaInc

yeah... should be easier now....

13. monroe17

Okay so I see you used the second one.. But, I'm confused how you got those values... ln(A)=4lnx+3x when A= x^4e^{3x}

14. monroe17

is it..ln(x^4)=4lnx

15. dpaInc

$$\large y=ln(\frac{x^4e^{3x}}{(2x-5)^{\frac{1}{2}}}) =ln(x^4e^{3x})-ln(2x-5)^{1/2}$$

16. dpaInc

that's the first step...

17. dpaInc

now.... $$\large ln(x^4e^{3x})=lnx^4+lne^{3x}=4lnx+3x$$ do u undestand this part?

18. dpaInc

ooo... sorry my bad... the second part is wrong....

19. monroe17

ohhh... ln(A^b)=b*ln(A) ln(x^4)=4*ln(x) and e&ln cancel out leaving 3x

20. dpaInc

$$\large ln(2x-5)^{1/2}=\frac{1}{2}ln(2x-5)$$ that's what it should have been...

21. dpaInc

and yes (to your last post)

22. dpaInc

lemme rewrite what i shouldve written....

23. dpaInc

$$\large y=ln(\frac{x^4e^{3x}}{(2x-5)^{\frac{1}{2}}}) = 4lnx + 3x - \frac{1}{2}\color {red}{ln}(2x^5-1)$$

24. monroe17

$\frac{ 4 }{ x }+3..$

25. dpaInc

yep... keep going...

26. monroe17

I need to do chain rule?

27. dpaInc

yes, chain rule: $$\large [lny]'=\frac{y'}{y}$$

28. dpaInc

so $$\large [ln(2x^5-1)]'=\frac{[2x^5-1]'}{2x^5-1}$$ simplify....

29. monroe17

$\frac{ 4 }{ x }+3-\frac{ 1 }{ 2 }\frac{ 10x^4 }{ 2x^5-1 } = \frac{ 4 }{ x }+3-\frac{ 5x^4 }{ 2x^5-1 }$

30. dpaInc

haha... looks good..:)

31. monroe17

why haha? did I do something wrong?

32. dpaInc

no... i was getting the wrong answers until at the last moment when u posted ur answer... it confirmed my last answer...

33. monroe17

oh lol :) you just discouraged me! >;[ no im jk ;) thank you!

34. dpaInc

oh wait...

35. dpaInc

haha... just kidding.... yw... :)

36. monroe17

haha! jerk ;D

37. dpaInc

:)

38. dpaInc

i forgot... nice work on your part... way to stick with the problem....

39. monroe17

what you mean? I always finish ;D