Derivative of Sin?

Question

What is the derivative of sin(x + (π/2))

Is it: cos (x + (π/2))? Or is there a chainrule involved?

Answer 1

Let u = x + π/2
du/dx = 1
y = sin u
dy/du = cos u
dy/du = cos (x + π/2)
dy/dx = (dy/du) (du/dx)
dy/dx = cos (x + π/2 )

Answer 2

Your answer is correct; the derivative w/r/t x of (x + (π/2)) is 1. Using the trig identity for cos (a + b) will simplify the result to -sin x.

Answer 3

cos (x + (π/2))

ok so it is sin(x+(pie/2))(1)

you take the deriv of the outside, keep the inside stuff and then the inside.

Answer 4

d/dx(sin(x + (π/2)) = cos (x + (π/2))

There is a chain rule involved, but the derivative of x + (π/2) is 1. cos (x + (π/2))*1 = cos (x + (π/2)).

By the way sin(x + (π/2)) = sin(x)cos(π/2) + cos(x)sin(π/2), and cos(π/2) = 1, and sin(π/2) = 0, so sin(x + π/2) = cos(x). cos (x + (π/2)) = -sin(x)

Answer 5

Chain Rule (f(g(x)) ' = f ' (g(x)) * g'(x) First permit f(x) = sin x and g(x) = e^x. Then f ' (x) = cos x and g ' (x) = e^x (sin (e^x)) ' = cos (e^x) * e^x Now permit f(x) = e^x and g(x) = sin x. Then f ' (x) = e^x and g ' (x) = cos x (e^(sin x)) ' = e^(sin x) * cos x answer: e^x cos (e^x) + cos x e^(sin x)