hooverst Group Title y=e^xlnx y=ln(3x^2) y=lnx/x^2+1 2 years ago 2 years ago

1. dpaInc Group Title

what's the question? which equation is nicest?

2. dpflan Group Title

What's the goal? Solve for x in this system of equations?

3. dpflan Group Title

heh, I like the first one ;)

4. dpaInc Group Title

me too! :)

5. hooverst Group Title

each one is a different question its says compute the derivatives of each function

6. dpflan Group Title

Oh, OK, are you familiar with (1.) product rule and (2.) quotient rule and the (3.) power rule?

7. hooverst Group Title

yeah a little

8. dpflan Group Title

(1.) $y=e^x*ln(x)$ (2.) $y=ln(3x^2)$ (3.) $y=\frac{ln(x)}{x^2}+1$

9. dpflan Group Title

And you know chain rule? a little, right? ;)

10. hooverst Group Title

the 1st equation is $y=e ^{xlnx}$

11. hooverst Group Title

yes

12. dpflan Group Title

OK, well, let's work with equation 2, for me that is the easiest one. We are differentiating with respect to y, right?

13. hooverst Group Title

yes

14. dpflan Group Title

Cool, $y=ln(3*x^2)$ We will use the Chain Rule here

15. dpflan Group Title

$\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}$ $u = 3*x^2$and $\frac{d*ln(u)}{du} = \frac{1}{u}$ $\frac{du}{dx} = 3*2*x^{2-1} = 6x$

16. dpflan Group Title

Yes, and to get du/dx we use the power rule

17. dpflan Group Title

$x^y = y*x^{y-1}$ is the power rule

18. dpflan Group Title

Can you finish solving the equation?

19. hooverst Group Title

ok cause when he explained it in class i did not understand it at all.

20. dpflan Group Title

All right, how is it now?

21. hooverst Group Title

ok so for the power rule what do i plug in or do I have to plug anything in

22. dpflan Group Title

Nah, you don't need to plug-in anything. So, what do you have so far? If you are writing it down, you can snap a picture with your phone if it has a camera, then post the picture here

23. dpflan Group Title

$\frac{d}{dx}(ln(3*x^2)) = \frac{d*ln(u)}{du}*\frac{du}{dx}$ $\frac{d*ln(u)}{du} = \frac{1}{u}$ $\frac{du}{dx} = 3*2*x^{2-1} = 6x$ $u=3*x^2$ So you have $\frac{1}{3*x^2} * 6x$ Simplify and you're done with that equation

24. hooverst Group Title

O ok I didn't know whether what I had was right but thats what I just got to.

25. dpflan Group Title

This maybe the hardest of the bunch because of the substitution, chain rule, and product rule

26. hooverst Group Title

yea It was

27. dpflan Group Title

Well, because substitution can be confusing because you introduce a new variable to differentiate on

28. dpflan Group Title

*or with respect to

29. hooverst Group Title

O ok

30. dpflan Group Title

Now (1.)$y=e^x * lnx$

31. Callisto Group Title

Excuse ... me.... are we finding dy/dx or dx/dy?

32. dpflan Group Title

dy/dx I believe

33. Callisto Group Title

34. dpflan Group Title

Heh, thanks, are you following along @Callisto ?

35. dpflan Group Title

$y=e^x* lnx$ Here we use the product rule. We can think of this equation as the product of 2 functions of x

36. dpflan Group Title

$(f*g)' = f'*g + f*g'$ So the derivative of the product of these functions, f and g Is the sum of the derivative of the first * the second + the first * the derivative of the second

37. Callisto Group Title

For the second one, I was thinking in this way.. $y=ln(3x^2)=ln3+ lnx^2 = ln3+2lnx$$\frac{dy}{dx}=\frac{d}{dx}2lnx = ...$

38. dpflan Group Title

Yes, that is easier! @hooverst If you did it that way you don't have do substitution

39. hooverst Group Title

Ok dpflan

40. dpflan Group Title

It can be really fun in math the find the simpler way of doing things, so you just use some intuition and understanding to make your work easier, simpler can be much more beautiful too

41. dpflan Group Title

*to find

42. dpflan Group Title

Nice one @Callisto @hooverst So, you think you can solve equation (1.) with the product rule?

43. dpaInc Group Title

hold on... i think @hooverst clarified that the first equation is: $$\large y=e^{xlnx}$$ , not $$\large y=e^{x}\cdot lnx$$

44. dpflan Group Title

Heh, you're right, man

45. hooverst Group Title

so callisto when you solved your equation for the second 1 what did you get for your anserw

46. dpflan Group Title

Thanks, getting a little carried away. Let me step away for a bit

47. dpaInc Group Title

ur doing fine....:)

48. Callisto Group Title

@hooverst What did you get ??

49. hooverst Group Title

is the anserw $y=\ln(3x^2)=1\div 3x^2$

50. Callisto Group Title

Not really... what is $$\frac{d}{dx}lnx$$ ?

51. hooverst Group Title

im not sure

52. dpaInc Group Title

are we still working on the second equation? @hooverst ???

53. dpaInc Group Title

because you can do this many different ways... but in the end, the derivative should be the same...

54. hooverst Group Title

yes but I was trying to figure out was 2lnx the anserw for #2 or is y=ln(3x^2)=1/3x^2 the anserw

55. dpaInc Group Title

answer as in the derivative? you and @dpflan , worked it out to $$\large \frac{6x}{3x^2}$$ simplified to....???

56. dpaInc Group Title

in the case with @Callisto , her method was to simplify the function first: $$\large y=ln(3x^2)=ln3+2lnx$$ so $$\large y'=[ln3]'+[2lnx]'=$$ ...???

57. dpaInc Group Title

and either way the derivative is the same...

58. Callisto Group Title

The key point is you need to know what $$\frac{d}{dx}lnx$$ is.

59. dpflan Group Title

Yes, that is a derivative you need to memorize, it will be quite useful

60. hooverst Group Title

Ok so many of you have said had your own opinion about the equation so which one is the right 1.

61. dpflan Group Title

Hehe, right, so, what is your opinion? ;p You can solve a problem many, many, different ways

62. dpflan Group Title

For you, how would you approach it now that you've seen how we would?

63. Callisto Group Title

I would say, none of us have given you the final answer. But we have given you some steps you need using different approaches. Though, the final answer will be the same.

64. hooverst Group Title

Ok Think the one Callisto gave me is more simple for #2

65. dpflan Group Title

Definitely, that was an awesome application of intuition

66. dpflan Group Title

well, and mathematical understanding

67. dpaInc Group Title

i guess it comes down to preference because i personally would've used u'/u ... the method used earlier...

68. Callisto Group Title

The step I left for you is to find $$\frac{d}{dx}lnx$$. Multiply the answer you get for that by 2. Then, it's done.

69. hooverst Group Title

ok

70. Callisto Group Title

May I ask you again- what is d/dx ( lnx) ?

71. hooverst Group Title

would you multiply it.

72. Callisto Group Title

Nope.... Hint: look it up in the pdf.: http://tutorial.math.lamar.edu/pdf/Calculus_Cheat_Sheet_Derivatives.pdf You can find the answer for d/dx (lnx) there.

73. hooverst Group Title

Ok is it d/dx(ln(x))=1/x

74. Callisto Group Title

Yes. dy/dx = d/dx (ln3 + lnx^2) = d/dx (2lnx) = 2 d/dx(lnx) = ...?

75. hooverst Group Title

is it 2/x or just 2x Idon't know if its right

76. Callisto Group Title

Which one do you think? (i) 2/x (ii) 2x

77. hooverst Group Title

2/x

78. Callisto Group Title

Yes. That's correct. Any questions?

79. hooverst Group Title

just 1 when we started simplifying the equation where did you get the 2 from?

80. Callisto Group Title

$lnx^a = a\ln x$

81. hooverst Group Title

Ok I get it. thanks Callisto

82. Callisto Group Title

Welcome.