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Finding the derivative of F (x) = [cos (INX)] ^ 2
This is a composite function, f (x) = x ^ 2g (x) = cosxh (x) = LNX, so f {G [H (x)]} = [cos (INX)] ^ 2. So first, we derive the square, then cos, and finally LN, so f '(x) = 2cos (LNX) * [cos (LNX)]' = 2cos (LNX) * [- sin (LNX)] * (LNX) '= 2cos (LNX) * [- sin (LNX)] * (1 / x)
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