how do i find the derivative of y=2sin8(x+pi/4)-12??

let u= 8(x+PI/4)

then u'= 8

y'=-2cos u * u'

y'=-16cos(8(x+PI/4))

thank you so much I can see where I went wrong now.=D

To find the derivative of the given function y = 2sin(8(x + π/4)) - 12, we can follow these steps:

Step 1: Apply the chain rule. The chain rule states that if we have a function of the form f(g(x)), then the derivative is given by f'(g(x)) * g'(x). In our case, g(x) = 8(x + π/4) and f(x) = 2sin(x).

Step 2: Find the derivative of g(x). Since g(x) = 8(x + π/4), the derivative of g(x) with respect to x (g'(x)) is simply 8.

Step 3: Find the derivative of f(x). The derivative of f(x) = 2sin(x) with respect to x is cos(x), by using the derivative property of sine function.

Step 4: Apply the chain rule. Multiply f'(g(x)) and g'(x). So, the derivative of y with respect to x is given by:

dy/dx = f'(g(x)) * g'(x)
= (2cos(g(x))) * g'(x)
= 2cos(g(x)) * 8
= 16cos(8(x + π/4))

Therefore, the derivative of y = 2sin(8(x + π/4)) - 12 with respect to x is dy/dx = 16cos(8(x + π/4)).