Question

h(x) = sin2 x + cos x 0 < x < 2π

Answer 1

hello, is this for calculas?

Answer 2

yes I need help with understanding derivatives for cofunctions.

Answer 3

Calculus I

Answer 4

ok r u looking for h'(x)?

Answer 5

Yes. So the concept here is finding derivatives. In this situation product rule wouldn't apply correct?

Answer 6

cos 2x + -sin x

Answer 7

would this be correct?

Answer 8

one sec

Answer 9

Nope, don't forget that you have to multiply the derivative of the substituted in the chain rule

Answer 10

nope

Answer 11

f'(x) = cos(2x) (2)
f'(x) = 2cos(2x)

Answer 12

same for cosx

Answer 13

Wait, is it sin^2(x) or sin(2x)

Answer 14

The cos x part is correct though

Answer 15

sin(2x)

Answer 16

Don't forget to multiply the derivative of 2x after applying the chain rule

Answer 17

so f'(x)= 2cos(2x)-sinx

Answer 18

For example, [sin(5x)]' = 5cos(5x)

Answer 19

oh gotcha. this is true

Answer 20

Yep, that's perfect

Answer 21

thanks a lot. This helped out a lot. Ok so now how to convert this into radians.

Answer 22

You don't have to convert this into radians, because this was never degrees

Answer 23

remember, x is a variable

Answer 24

yes true. I need to find the critical numbers though and that requires conversion. Correct?

Answer 25

What are you required to find?

Answer 26

\[\frac{ d }{dx}\sin(2x)+\cos(x),(0<x<2\pi)\rightarrow 2\cos(2x)-\sin(x)!\]

Answer 27

Don't put the exclamation mark at the end, it is ambiguous to factorial

Answer 28

lol

Answer 29

listen to @kc_kennylau

Answer 30

there are two critical numbers for this equation between 0<theta<2pi

Answer 31

Set this to zero

Answer 32

That means solve h'(x)=0

Answer 33

gotcha. Yes I remember now. Well thanks again.

Answer 34

no problem

Answer 35

You know how to solve the equation?

Answer 36

https://www.khanacademy.org/math/differential-calculus/derivative_applications/critical_points_graphing/v/minima--maxima-and-critical-points

Answer 37

This should be helpful

Answer 38

I love khanacademy

Answer 39

GTG, HAPPY STUDYING!!!!!