Question

suppose f(-2)=0, F(0)=pi, f(2)=2pi, and f'(x)=sqroot(4-x^2)
if h(x)=f(2sinx) what is h'(0) i got 2cospi

what i did was i took the derivative of 2sinx which is 2cosx and then it says the output value for x is pi so my answer is 2cospi which i think is wrong

Answer 1

why would it be product rule

Answer 2

i thought 2sinx was 1 term

Answer 3

where did the f(x) come from?

Answer 4

oh i think i know what you are talking, it wouldnt it be the chain rule then

Answer 5

i viewed it as function f of x at 2sinx

Answer 6

im given what i wrote so your guess is as good as mine :D

Answer 7

crap
I mean h'(0) = pi

Answer 8

i know i hate it

Answer 9

i will most deffinatly ask that question but you have helped me a lot

Answer 10

h(x)=f(2sinx)

h(x) = f(x) when x is equal to 2sin(x)

Answer 11

its not the composition of two functions sorry

Answer 12

My brain has really been failing me lately ugh

Here is my try at this problem but I'm not sure if I'm right
h'(x) =f'(2sin(x))
so

h'(x) = (4-(2sin(x)')^2)^(1/2)'

h'(x) = (4-(2cos(x))^(2))^(1/2)'

h'(x) = (4+4cos(x)sin(x))^(1/2)'

h'(x) = 1/2(4+4cos(x)sin(x)) * (-4sin(x)sin(x) + 4cos(x)cos(x))

h'(0) = (1/2(4)) * (4)

h'(0) = 4/8 = 2

Answer 13

where did that sinx come from?

Answer 14

chain rule?

Answer 15

Here is my logic

if h(x) = f(2sinx)

f(2sin(x)) = h(x) than f(2sin(x)) rearranges in a way that makes it equal to h(x) so if you take the derivative of 2sin(x) and place it into the know f'(x) function you should end up with a derivative equal to h(x)

Answer 16

I don't know if that makes sense might be failed logic probably best you ask your ta or check your notes

Answer 17

its sound logic to me

Answer 18

I wouldn't take my word for it but that is just my reasoning and how I would answer it on a test if I had enough time lol after reading it incorrectly

Answer 19

when you took the derivative did you subtract 1 from the power?

Answer 20

I think I'm wrong actually I'm pretty sure im wrong

Answer 21

Just ask your profs :L