Question

I think I got it but I need someone to verify this exercise for me:

Look for the first and second derivative of:
f(x) = x + (2/(sin(x))) , (-pi, pi)

f(x) = x + 2csc(x) , (-pi, pi) <<< Rewrite

f'(x) = (2)(-csc(x)tan(x) + (0)(csc(x)) << 14 years ago

Answer 1

you dont really need to differentiate constant
(0)(csc(x)) <<<Differentiate

Answer 2

Yeah, but if you use the product rule, you get:
f'(x) = 1 - 2csc(x)tan(x) , (-pi, pi) <<< First Derivative

or not?

Thanks

Answer 3

i meant in the first ...

Answer 4

f'(x) = 1 + (2)(-csc(x)tan(x)

Answer 5

yeah that's what I got in the first derivative

Answer 6

Now what's the second derivative?

Answer 7

1 goes to zero

Answer 8

yeah, and what happens to the other part -2csc(x)tan(x) ?

Answer 9

you know how to use product rule??

Answer 10

yes

Answer 11

But we have 3 factors there, -2csc(x)tan(x), so how...?

Answer 12

and your derivative is also wrong .. i guess
http://www.wolframalpha.com/input/?i=d%28csc+x%29%2Fdx

Answer 13

because the derivative of cscx is -cscx cot x

Answer 14

you have tan up there

(2)(-csc(x)tan(x) + (0)(csc(x)) <<<Differentiate

Answer 15

yeah I see it now, sorry :L

Answer 16

hey i got zero. so that might be the answer.!

Answer 17

well wolftram shows that the 2nd derivative is :

http://www.wolframalpha.com/input/?i=d%281+-+2cot%28x%29csc%28x%29%29%2Fdx

so....

Answer 18

i dont think so. maybe ur right

Answer 19

No, wolftram is right :)

Thanks everyone, I understand now :)