To find the derivative of cos(2x) using first principles

 To find the derivative of cos(2x) using first principles, we start with the definition of the derivative:

f'(x) = lim h→0 [f(x+h) - f(x)] / h

In this case, f(x) = cos(2x), so we have:

f'(x) = lim h→0 [cos(2(x+h)) - cos(2x)] / h

Next, we use the trigonometric identity:

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)

to rewrite the numerator in terms of cosines and sines:

cos(2(x+h)) - cos(2x) = [cos(2x)cos(2h) - sin(2x)sin(2h)] - cos(2x)

= cos(2x)(cos(2h) - 1) - sin(2x)sin(2h)

= -2sin^2(x)sin^2(h) / h

where we have used the double angle formula for cos(2h) and the identity sin(2x) = 2sin(x)cos(x).

Substituting this expression back into the definition of the derivative, we get:

f'(x) = lim h→0 [-2sin^2(x)sin^2(h) / h] = -2sin^2(x) lim h→0 [sin(h) / h]

We recognize the limit in brackets as the definition of the derivative of sin(x) at x=0, which is equal to 1. Therefore, we have:

f'(x) = -2sin^2(x)

So the derivative of cos(2x) with respect to x, evaluated using first principles, is:

d/dx [cos(2x)] = -2sin^2(x)