Question

derive cos^2(x)+sin(x). Then find the critical numbers.

Answer 1

Derive or Derivate ?

Answer 2

Sorry, derivative. I had a late night and just stumbled upon this website.

Answer 3

\[f'(x)=2*\cos(x)*-\sin(x)+\cos(x)\]

Answer 4

Wait, how? Thanks though...

Answer 5

do you know the chain rule?

Answer 6

Yes, I got {(-sinx*cosx)+(cosx*-sinx)} + cosx.

Answer 7

what you calculated is the same as what I wrote.
:)

Answer 8

yes

Answer 9

btw, did you use the product rule on cos^2(x)
i.e. write it as cos(x) * cos(x) ?

you could also use the "power rule"
d u^2 = 2 u du
or in this case
d cos^2 x = 2 cos(x) d cos(x) = 2 cos(x) (-sin x)
or -2 cos(x) sin(x)

Answer 10

But I don't know how to solve once set to zero..

Answer 11

hint:
\[Cos(x)*(-2*Sin(x)+1)=0\]

Answer 12

-2sinx cosx+ cosx= 0

if we used a= sin x and b= cos x (just renaming ) this is the same problem as
-2 ab + b =0
factor out b:
b(-2a+1) = 0
either b is 0 or -2a+1 is 0

or, using the originals
cos x = 0 or -2 sin x + 1 = 0

Answer 13

cos x is zero at pi/2 , pi/2+ pi, pi/2 + 2pi , etc
in general
\[ \frac{\pi}{2} + n \pi \]

Answer 14

so the pi's would cancel, and 2 would be my answer?

Answer 15

no.

the cosine "goes forever" and it crosses the x-axis many times
http://www.mathsisfun.com/algebra/trig-sin-cos-tan-graphs.html

the "zeros" occur at pi/2 (90 degrees) and repeat every pi radians (every 180 deg) in both directions.