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1. Could anyone please show me how to work this out as I cannot get the correct answer?

If f(x) = x sin^2 x, find f ' (pie/6).

Thanks
2. What did you get for f'(x)?
3. f ' (x) = x(2sinxcosx) + sin^2 x

2sinxcosx = sin2x

f ' (x) = xsin2x + sin^2 x
4. That's correct. What did you get when you set ?
5. now just substitute x = pi/6
6. The book says that the answer is 1/4 ( 1+pie/root3 )

But the closest that I can get is pie/6 (2root3/4) + 1/4

I know where the 1 comes from, but its the pie/root3 that I cannot get.
7. f(x)= sin^2x = (sinx)^2

f'(x)= 2sinx . cosx = 2sinxcosx (using the chain rule)

now just sub pie/6 into x

pretty sure that is right however haven't done maths since my last exam in june
8. (Original post by katbro20)
The book says that the answer is 1/4 ( 1+pie/root3 )

But the closest that I can get is pie/6 (2root3/4) + 1/4
Aren't these equivalent? Try rationalising the denominator of the book's answer and see if you can show they're the same.
9. soory, didn't see the x at the beginning
10. f(x)=xsin^2x u= x u'=1 v=sin^2x = (sinx)^2 v'= 2sinxcosx

f'(x)= x(2sinxcosx) + sin^2x

2sinxcosx = sin2x (double angle)

therefore, f'(x) = xsin2x + sin^2x (now just sub pie/6 in to x)
12. Thats where I got up to.

The problem that I am having is substituting in the pie/6
13. do it one at a time, don't enter it all into the calculator at once e.g. find sin2(pie/6), then multiply that answer by pie/6 etc
14. no calculator. The answer is given as 1/4 (1 + pie/root3)
15. (Original post by katbro20)
Thats where I got up to.

The problem that I am having is substituting in the pie/6
16. Am I missing something obvious?
17. don't know how they got that but its equivalent to the decimal value (0.7034), so they won't mark you down if like you done that in the exams
18. Yeah, their answer is an a slightly unusual form. Not too hard to show they're equivalent though.
19. OK. Lets see if I can do this.

If pie/6 (2root3/4) + 1/4 = 1/4 (1 + pie/root3)

Then pieroot3/12 + 1/4 must = 1/4 + pie/4root3

pieroot3/12 = pie/4root3

pieroot3/3 = pie/root3 reverse of rationalising

which gives pie/root3 = pie/root3

Unless my logic has gone wrong somewhere, they are equal (thanks Calira), therefore, I must be able to change my version into their version some how.
20. Thanks everyone. Finally managed to do it.

Now it is do, I cannot believe that I missed taking out the common factor from the first part of the function.

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