What is the derivative of y=sin(x+c)?

Answer 1

y'=cos(x+c)
where c is constant
the rule is
if
y=sin(s) where s is a function of x
so
dy/dx = cos(s) * ds/dx
here in ur example s=x+c , ds/dx =1
so the derivative will be cos(x+c)*1

Answer 2

dy/dx sin u = (cos u)(du/dx)

in the problem, u = ( x + c )

du/dx = 1
because the derivative of x is one and c, a constant is zero.


y' = [cos ( x + c )][ 1 ]
y' = cos ( x + c )

Answer 3

find derivative of da sin term 1st irrespective of da angle given ie cos(x+c).now multiply wid da derivative of da angle ie(x+c)=1.
therefore - dy/dx=cos(x+c)*1=cos(x+c)

Answer 4

use the chain rule:
let u = x + c
then d/du sin(u) = cos(u)

d/dx (x + c) = 1

y' = cos(x + c) * 1 = cos(x+c) <== answer

Answer 5

y=sin(x+c)
Then dy/dx=cos(x+c) . d/dx (x+c)
=cos(x+c) *1
=cos(x+c)

Good luck!

Answer 6

dy/dx = cos (x + c)