Question

Calc 1 please help.

Answer 1

Have a specific question bear? :)

Answer 2

yes i am posting in momentarily

Answer 3

oops

Answer 4

is there a better way to post than word

Answer 5

use Quotient rule:\[(\frac{f}{g})'=\frac{f'g-g'f}{g^2}\]

Answer 6

did that and got a very messy answer (4cosx+4cosxsinx^2-8sinx^2-4sinxcosx)/(1+sinx^2)^2

Answer 7

\[\frac{ (1+\sin^2x)(4cosx)-(4sinx)(2sinx(cosx)) }{ 1+\sin^2x }\]

Answer 8

then i set it equal to zero. so 4cosx+4cosxsinx^2-8sinx^2-4sinxcosx=0

Answer 9

you forgot to square the bottom

Answer 10

oops :D

Answer 11

lol

Answer 12

but now how do i find the answer(s). I set it equal to 0

Answer 13

they are get the numerator to \[4\cos^3(x) \]

Answer 14

Hmmm your numerator is 4cos^3(x)? Hmmmm