## Carissa15 one year ago Hi, I have a few derivative problems if anyone is able to help please?

1. triciaal

|dw:1438832647688:dw|

2. Carissa15

1) $f(x)=\tan(x^3)$ I have then used the chain rule as follows (dx of outside) x (inside) x (derivative of inside)

3. triciaal

use u substitution

4. UsukiDoll

but we're just taking derivatives, not anti-derivatives where u-sub thrives. o_O

5. Carissa15

Which I get$=\sec(x^3) * 3x^2$

6. UsukiDoll

isn't derivative of tan (x) sec^2x though?

7. UsukiDoll

the derivative of x^3 is correct.. it does become 3x^2 and is written outside

8. Carissa15

So my method is correct but should be $f \prime (x)=3x^2 \sec^2(x^3)$?

9. UsukiDoll

yeah

10. Carissa15

awesome, thank you. I have a couple more but not sure where to start as they include $\ln and$

11. UsukiDoll

trig derivatives sin (x) -> cos(x) cos (x) -> -sin(x) tan (x) -> sec^2(x) csc(x) - > -csc(x)cot(x) sec(x) -> secxtanx cot(x) = -csc^2x

12. UsukiDoll

yes that table needs to be memorized just like all the derivative rules out there x-X!

13. Carissa15

do you know what the rules are for e? another question is $f(x)=e^xsinx$

14. Carissa15

I thought it would be $f \prime (x)=e \cos(x)$

15. UsukiDoll

yeah... usually for e^x the exponent part for the e is left alone so for example the derivative of e^x is just e^x due to the fact that the derivative of x is 1, but nobody writes the 1 so derivative of x is one , but the exponent part that's attached to the e stays e^{x}(1)

16. UsukiDoll

another example |dw:1438833198807:dw|

17. UsukiDoll

the derivative of x^2 is written in front of the e but the exponent part where x^2 stays is left alone

18. Carissa15

Oh ok, sorry just read your response. Great thanks. So in the second question it is only the sin to cos that changes?

19. UsukiDoll

yes because we are using product rule on that one $\LARGE f(x)=e^xsinx$ $\LARGE f'(x)=e^x(cos(x))+sin(x)e^{x}$

20. UsukiDoll

and the derivative of that x is just one so nothing drastic happens. . .

21. UsukiDoll

I left the e^x alone first and took the derivative of sinx + I left sin x alone and took the derivative of e^x derivative of e^x in that case is just e^x(1) or just written as e^x

22. Carissa15

great thank you, could it also be$e^x(\cos(x))+e^x(\sin(x))$

23. UsukiDoll

yeah... we could factor the e^x to make it nicer, but it's fine the way it is.

24. UsukiDoll

order doesn't really matter that much.

25. Carissa15

cool, makes sense :) Thanks

26. Carissa15

I also have $f(x)=\frac{ \ln x }{ x }$ I know that the dx of $\ln (x)$ is $\frac{ 1 }{ x }$ but not sure what to then do about the x underneath?

27. UsukiDoll

the derivative of x is one so the derivative of ln(x) is just written as 1/x

28. UsukiDoll

maybe quotient rule will work if we had to do $\LARGE f(x)=\frac{ \ln x }{ x }$

29. UsukiDoll

so the bottom gets squared then leave the second part (The denominator x alone) deal with ln x the derivative of ln x is 1/x(1) but again since the derivative of x is just 1 we don't have to write that extra stuff.. just 1/x will be good enough. - leave the first part (the numerator ln x alone ) deal with the x the derivative of x is just 1

30. UsukiDoll

$\LARGE f(x)=\frac{ \ln x }{ x }$ $\LARGE f'(x)=\frac{ (x) \frac{1}{x}- ln(x)(1) }{ x^2 }$ $\LARGE f'(x)=\frac{\frac{x}{x}- ln(x)(1) }{ x^2 }$ $\LARGE f'(x)=\frac{1- ln(x) }{ x^2 }$

31. Carissa15

(f’ g − g’ f )/g2 so I get$\frac{ 1 }{ x } * x -1 * lnx$ over x^2

32. UsukiDoll

yeah we can cancel the x/x and we don't have to write the 1 for the lnx

33. Carissa15

ok, cool. thanks

34. Carissa15

So for $f(x)=e^\sin x$

35. Carissa15

as the sin x is an exponent of e it remains untouched?

36. UsukiDoll

the latex broke...

37. UsukiDoll

$\LARGE f(x) = e^{sinx}$ like this?

38. Carissa15

oh, yes

39. UsukiDoll

so take the derivative of sin(x) and put it in front of the e. leave the exponent part of e which is at e^{sinx} alone.

40. Carissa15

oh, yeah i forgot that we still multiply but do not change the original as well. Cool. Thank you. I have one more question.

41. Carissa15

$f(x)=\ln(\sin^2x)$

42. Carissa15

would it be$\frac{ 1 }{ x } *\cos^2$

43. UsukiDoll

wow... ok to make it easier we can re-write that $\LARGE f(x)=\ln((sinx)^2)$

44. UsukiDoll

then we know that since we have a log it will be 1/(sinx)^2 but then we have to take the derivative of (sinx)^2

45. UsukiDoll

$\LARGE f(x)=\ln((sinx)^2) \rightarrow \frac{1}{(sinx)^2}2(sinx)\cos(x)$

46. UsukiDoll

$\LARGE f'(x) = \frac{2\cos(x)}{\sin(x)}$

47. UsukiDoll

I had one sin x on the top but 2 sin x 's on the bottom.. one of them cancels..

48. Carissa15

great, that makes much more sense now.

49. UsukiDoll

yeah but here's the thing.. .when I checked on wolfram it used a trig identity see that 2sinxcosx ? sin2x = 2sinxcosx

50. Carissa15

oh

51. UsukiDoll

$\LARGE f'(x)=\ln((sinx)^2) \rightarrow \frac{1}{(sinx)^2}\sin2x$ interesting... we need trig identities. according to mr.wolfram the derivative is 2 cot(x)

52. Carissa15

hmmm. I got this on mathway 2csc2(x)cos(x)sin(x)

53. Carissa15

but not sure how it got there....

54. UsukiDoll

ok hold up let's take the (sinx)^2 out for a bit let let a = (sinx)^2 so we just have ln a for now

55. UsukiDoll

$\LARGE f(x) = \ln(a) \rightarrow f'(x) = \frac{1}{a}$

56. UsukiDoll

so if we have $\LARGE a = (sinx)^2$ $\LARGE a' = 2(sinx)(cosx)$

57. UsukiDoll

however there is a trig identity so that 2sinxcosx is replaced with sin2x

58. UsukiDoll

OMG. ok let's use log rules on this beast.

59. UsukiDoll

exponent rule is needed so... $\LARGE f(x)=\ln((sinx)^2)$ becomes $\LARGE f(x)=2\ln((sinx))$

60. UsukiDoll

and then use product rule... YEAH I SEE IT NOW!

61. UsukiDoll

we just need one part of the product rule because 2 is a constant and all constants have the derivative of 0

62. UsukiDoll

$\LARGE f(x)=2\ln((sinx))$ $\LARGE f'(x)=2(\frac{1}{sinx})(cosx)+ \ln(sinx)(0)$

63. UsukiDoll

$\LARGE f'(x)=2(\frac{cosx}{sinx}) \rightarrow 2\cot(x)$

64. UsukiDoll

sh.it I had that earlier ... just forgot to take the identity of cosx/sinx

65. UsukiDoll

12 minutes ago I had this $\LARGE f'(x) = \frac{2\cos(x)}{\sin(x)}$

66. UsukiDoll

cosx/sinx = cot x

67. Carissa15

I have not come across trig identities yet, another set of rules for specific trig functions?

68. Carissa15

oh ok

69. UsukiDoll

trigonometry comes before calculus... so eventually by now all the trig identities have to be mastered and memorized... I just overlooked that one part.

70. UsukiDoll

$\LARGE f'(x) = 2 \cot(x)$

71. UsukiDoll

so I didn't have to use sin2x = 2sinx cosx after all... just had to cancel one of the sinx and then use cosx/sinx = cotx and that's what wolfram got too.

72. Carissa15

awesome, I will have a closer look at trigonometry. Thank you so much for all of your help. Has been amazing!

73. UsukiDoll

yeah... it wasn't too bad... except for the last one when I overlooked something... XD! and then taking the identity of 2sinxcosx made it worse. XD

74. Carissa15

All good, was so lost without your help. Thanks again, I will close this now :)

75. UsukiDoll

ok :)