Does anyone know how to find the derivative to this problem

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

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

First, the derivative of the whole thing is just the difference in the derivatives of each part, so we need to find the derivative of cos^2x and the derivative of sin^2x. This can be done by the chain rule:

Derivative of cos^2x = 2cosx * -sinx
Derivative of sin^2x = 2sinx * cosx

So, putting this back together, the derivative is:

-2cosxsinx - 2sinxcosx, which is:

-4cosxsinx

∵ ƒ(x) = cos² x - sin² x = cos 2x

∴ by Chain Rule,

ƒ'(x) = d/dx[ cos (2x) ]

. . . .= [ -sin (2x) ] • d/dx( 2x )

. . . .= ( -sin 2x ) • 2( 1 )

. . . .= -2 sin 2x .............................. Ans.

.OR = -2 ( 2 sin x cos x )

. . . .= -4 sin x cos x ....................... Ans.
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You should be able to do this.
1 = s²+c² (s=sine(x), c=cosine(x))
so s² = 1-c²
so c²-s² = c² -(1-c²) = 2c²-1
c² = c*c
d(ab) = b*da/dx+a*db/dx for any a(x) and b(x) by the chain rule
so dc² = 2c(dc/dx) and you should have memorized what dc/dx is.

yes