Question

Derivative of
y=C(1)sin(3x+C(2))

is the answer to this 3C(1)cos(3x+C(2)) or is it 1/3C(1)cos(3x+C(2))???

Answer 1

while you've differentiated this, u'd use chain rule to get
d/dx(3x+C(2)) = 3
so, that would get multiplied to it, and not divided.
and its 3C(1)cos(3x+C(2))

Answer 2

thank you.

Answer 3

welcome ^_^

Answer 4

okay wait, just a quick question. I don't know why i am confusing myself. But if I do u=3x+C2 then du=3x+C2 dx.... I though you solve for dx and you have du/3x+C2 = dx
But this is how you would write this correctly??

dy=C1sin(u) du

Answer 5

and the du = 3x+C2 dx... and thats how you get the 3??
I think I was just doing it backwards and confusing myself wtith the notation.

Answer 6

just a sec, if u=3x+C1
then du = 3dx
because C1 is constant(when you differentiate w.r.t x) and derivative of constant =0

Answer 7

so, if you've put u=3x+C2 in y=C(1)sin(3x+C(2))
du = 3 dx
dy = c1 sin u du = 3c1 sin (3x+c2).

Answer 8

dy = c1 sin u du = 3c1 sin (3x+c2).dx

Answer 9

when you use U substitution for integration it is the same process right? and still take the derivative the same way?

Answer 10

yes, absolutely :)

Answer 11

okay. thanks again, much appreciated.

Answer 12

welcome again ^_^

Answer 13

integrating cos 2x, would give me 2sin2x right?

Answer 14

no, see this \(\int \cos 2x dx \\ put \: 2x =u , dx =du/2 \\ \int \cos u du/2 =1/2 (\sin u)+c =(1/2)sin 2x+c\)
got this ?

Answer 15

okay, thats how i did it, but i guess i am getting the rules confused between derivative and integration when using the chain rule.

Answer 16

anyways, thats again for the 3rd time.

Answer 17

are you starting integration ?

Answer 18

if yes, and when you get time, you can go through this :
http://openstudy.com/users/hartnn#/updates/50960518e4b0d0275a3ccfba