A math problem.... PLZ HELP!!?

Question

assum f(x) = sin( cos (sin) )
And f'(x◦) = 0 , and -л/2 < x < л/2

then what is x◦ ??


p.s: if u have any problem to understand the question please let me know... thanks a lot!

I do mean derivative!

Answer 1

f(x) = sin(cos(sin(x))) so
f'(x) = [cos(cos(sin(x)))]*[(-sin(sin(x)))]*[cos(x)]
so at least one of these three terms must be 0. By inspection, only the 2'nd term can be zero over the given domain and that occurs at x = 0 when
-sin(sin(0)) = -sin(0) = 0.

HTH ☺

Doug

Answer 2

x=0

Answer 3

f(x) = sin( cos (sin) )

is it f(x) = sin( cos (sin x) )???

if so then u have to use chain rule and get the general solution and get values that lie in -л/2 < x < л/2
applying chain rule
f'(x)=cos(cos (sin x)) *(-sin(sin x)) *cosx=0

remember d/dx cosx=-sinx and d/dx sinx=cosx

cos(cos (sin x)) *(-sin(sin x)) *cosx=0

so either cos(cos (sin x)) =0....#1 OR -sin(sin x)=0 ...#2 OR cosx=0...#3

for #3,x=pi/2 + or - k2pi where k is an element of integers
or x=3pi/2 + or - k2pi or if since -л/2 < x < л/2,there is no x in (-л/2,л/2) unless if the endpoints where included.for #2 and #1, u follow a simlar method but with a little bit of thinking on your part

Answer 4

do you mean the derivative?
it's -cos(cos(sin(x)) * sin(sin(x)) * cos(x)
you just apply chain rule
Now, that is 0 when cos(cos(sin(x)) = 0 or sin(sin(x)) =0 or
cos(x) = 0 . The latter and the first dont have solution in the given interval.sin(sin(x)) =0 , means sin (x) = 0 and this implies x = 0

Answer 5

trying to solve you will have

sin(x)=+-pi/2>1
then x is not real but complex one!

Answer 6

eeerrrrr?????