Is it y=cos5x or is it y=5cos5x or is it something else?
2007-11-22 17:38:08 · 9 answers · asked by ~*~ 1 in Science & Mathematics ➔ Mathematics
y = sin(5x)
Let u=5x
du/dx = 5
y = sin(u)
dy/du = cos(u) = cos(5x)
dy/dx = (dy/du)(du/dx)
= cos(5x) * 5
= 5cos(5x)
2007-11-22 17:41:46 · answer #1 · answered by gudspeling 7 · 2⤊ 2⤋
You use chain rule for this problem. You have to take the derivative of the inside, so the answer is most likely the second one.
y = sin5x
y' = 5cos5x
2007-11-22 18:53:52 · answer #2 · answered by jayj 3 · 0⤊ 0⤋
5 cos5x dats it
2007-11-22 19:11:59 · answer #3 · answered by kicks 1 · 0⤊ 2⤋
yup it's the second one, y = 5cos5x
Just remember chain rule, so you always have to remember to take the derivative of the inside of the trig function as well (in this case the 5x).
2007-11-22 17:42:19 · answer #4 · answered by Anonymous · 3⤊ 2⤋
y=sin 5x
So,dy/dx=cos 5x .d/dx 5x
=5 cos 5x
Your 2nd answer is correct.
Good luck!
2007-11-22 17:51:28 · answer #5 · answered by Anonymous · 0⤊ 2⤋
5cos5x
2007-11-22 17:41:25 · answer #6 · answered by Siwelttap 3 · 0⤊ 2⤋
y=sin(kx) with k=constant
=>y'=dy/dx=kcos(kx)
k=5=>y'=5cos5x!
y=cos(kx) with k=constant
=>y'=dy/dx=(-k)sin(kx)
2007-11-22 20:30:08 · answer #7 · answered by kami 5 · 0⤊ 2⤋
y'=5 cos (5x)
2007-11-22 20:14:34 · answer #8 · answered by blare_out 2 · 0⤊ 2⤋
y = sin (5x)
let u = 5x
du/dx = 5
y = sin u
dy/du = cos u
(dy/dx) = (du/dx)(dy/du)
dy/dx = 5 cos u
dy/dx = 5 cos 5x
2007-11-23 03:17:18 · answer #9 · answered by Como 7 · 0⤊ 3⤋