# How to find the antiderivative of cot(x)

1. Jan 11, 2009

### sonofjohn

5) As you can see ln(sin(1)) +1/ln.5 does not yield one of the above answers. Can someone tell me where I went wrong?

2. Jan 11, 2009

### math8

Re: antiderivative(cotx)dx

I think you made a mistake when finding C.
C should be equal to 1-ln(.5) not 1/ln(.5).
The answer should be the last one (E).

3. Jan 11, 2009

### sonofjohn

Re: antiderivative(cotx)dx

Thanks for the help, didn't see that.

4. Jan 11, 2009

### sonofjohn

Re: antiderivative(cotx)dx

Does anyone have an idea as to how to figure 6 out? I don't believe I have ever tried one of those problems.

5. Jan 11, 2009

### math8

Re: antiderivative(cotx)dx

You should try finding out what is the second derivative of [f^2(x)].
When you're done, you just plug in the number 3 and it is easy to find out what's the answer.

6. Jan 11, 2009

### sonofjohn

Re: antiderivative(cotx)dx

isn't the second derivative of f^2x = 2?

7. Jan 11, 2009

### math8

Re: antiderivative(cotx)dx

No, consider [f^2(x)]=[f(x)]^2
you know that (u^m)'=m (u^(m-1)) u' right?
So it's the same thing for u=f(x) and m=2.

Can you try it and see what you get when you do it twice?

8. Jan 11, 2009

### NoMoreExams

Re: antiderivative(cotx)dx

Certainly not. Let $$f(x) = x^3$$, then $$f^{2}(x) = x^6$$. Still think your rule works?

9. Jan 11, 2009

### sonofjohn

Re: antiderivative(cotx)dx

I can't see how to take the second derivative of f^2(x). Am I supposed to use the previous f(3) = 2 and ect for this? I have never actually learned how.

10. Jan 11, 2009

### NoMoreExams

Re: antiderivative(cotx)dx

Use the chain rule, if I told you to differentiate [sin(x)]^2 what would you do?

11. Jan 11, 2009

### sonofjohn

Re: antiderivative(cotx)dx

f(x)^2
2f(x)
and then what?

12. Jan 11, 2009

### NoMoreExams

Re: antiderivative(cotx)dx

No, that's not it. Try again :) You are saying if we define f(x) = sin(x) then f^(2)(x) = sin^2(x) and you are saying derivative of f^(2)(x) is 2*sin(x)? You know that's not true...

13. Jan 11, 2009

### sonofjohn

Re: antiderivative(cotx)dx

could I possibly multiply them and get f(2x^2)?

14. Jan 11, 2009

### math8

Re: antiderivative(cotx)dx

[f(x)^2]'= 2f(x)*f '(x)
you're not using the chain rule correctly. Don't forget the f '(x) at the end :)

Now can you find out what is [2f(x)*f '(x)] ' using the product rule?

15. Jan 11, 2009

### NoMoreExams

Re: antiderivative(cotx)dx

Where are you getting 2x^2 from? Think of f^2(x) as f(x)*f(x). What's the derivative of that. That uses the product rule instead of the chain rule, if you have trouble remembering that one.

16. Jan 11, 2009

### sonofjohn

Re: antiderivative(cotx)dx

2f'(x) + f"(x)2f(x) would that be it and then just number plug?

17. Jan 11, 2009

### NoMoreExams

Re: antiderivative(cotx)dx

No! Why is the first term 2f'(x)?

18. Jan 11, 2009

### sonofjohn

Re: antiderivative(cotx)dx

it should be 2f'(x)f'(x) + f"(x)2f(x) = 42. I messed up the chain rule again :(

Last edited: Jan 11, 2009
19. Jan 11, 2009

### sonofjohn

Re: antiderivative(cotx)dx

Now 7 is another problem I have yet to face. I see that t is in seconds so I think I should multiply t by 60 to convert it to minutes. Should I then plug 1 in for y and take the derivative?

20. Jan 11, 2009

### NoMoreExams

Re: antiderivative(cotx)dx

Actually that's the product rule.

If $$\frac{dy}{dt} = ky$$ Then y = ? Solve this separable DE.