Derivative of Trig Functions: Solving for f'(x) and f(∏/6)

[domyy]

Homework Statement

x³ - sin 2x

Find f'(∏/6)

The Attempt at a Solution

f'(x) = 3x² - 2 cos 2x
f(∏/6) = 2700 - 2 [ (√3/2) ] ---> from 2 [ cos(∏/6)]

answer: 2700 - √3

My book has the answer as (∏² - 12)/12

 

[lionely]

Work in radians so...

3(pi^2/36) - 2cos2(pi/6)

(pi^2/12) - 1 = (pi^2-12)/12

I hope I helped :S

 

[domyy]

But is it wrong if I convert ∏/6 = 30 and work with that?

 

[lionely]

Umm when I put (pi/6) into the gradient function 3x^2 - 2 cos 2x

I get 3(pi/6)^2 which is 3 (pi^2/36) minus 2cos2(pi/6)

2 times pi/6 is pi/3 and the cos of pi/3 is 1/2 and 1/2 times 2 is 1.

so the whole thing ends up being (pi^2/12)- 1
1= 12/12
so the last line is (pi^2)-(12)/12

 

[domyy]

But isn't (∏)^2/12 = 2700 and 12/12 = 1

Wouldn't it be sort of equivalent to 2700 - sqrt of 3 (1.73) ?

That's what I am trying to understand. What's the difference if I work with 30 instead of ∏/6 ? Is it wrong? =/

 

[lionely]

Wait.. how did you get 2700?

 

[domyy]

Oh I was thinking of ∏ = 180.
Is it wrong? =/

 

[lionely]

ohhhhh lol yes it's wrong don't think of it like that. In questions
you're either working in radians or degrees . In this case it is radians.

So so pi/6 is not 180/6 , if it was degrees then it would be 30 degrees.

 

[fortissimo]

Oh I was thinking of ∏ = 180.
Is it wrong? =/

It is very, very wrong. Pi radians is 180 degrees yes. But the number pi is not equal to the number 180.

 

[domyy]

nr. pi = 3.14, right?
Good you clarified that. Now I won't make the same mistake :)
180 and 3.14 were sort of mixed in mind.
Thank you so much! =D