Determinare la derivata rispetto a x di:
(a)  (x+1)(x-1)	(b)  (√x+1)(√x-1)	(c)  (√x+1)(√x-1)(x2-1)

Abbreviamo D_x con D.
(a)  D((x+1)(x−1)) = D(x+1)(x−1) + (x+1)D(x−1) = 1·(x−1) + (x+1)·1 = 2x
alternativa:  D((x+1)(x−1)) = D(x²−1) = D(x²) − D(1) = 2x − 0 = 2x

(b)  D((√x+1)(√x−1)) = D(√x+1)(√x−1) + (√x+1)D(√x−1) = 1/(2√x)(√x−1) + (√x+1)·1/(2√x) = 1/(2√x)(√x−1+√x+1) = 1/(2√x)(2√x) = 1
alternativa:  D((√x+1)(√x−1)) = D((√x)² − 1) = D(x − 1) = 1

(c)  D((√x+1)(√x−1)(x²−1)) = D((√x)² − 1)(x²−1)) = D((x−1)(x²−1)) = D(x−1)(x²−1) + (x−1)D(x²−1) = (x²−1) + 2x(x−1) = 3x²−2x−1
alternativa:  D((√x+1)(√x−1)(x²−1)) = D((√x)² − 1)(x²−1)) = D((x−1)(x²−1)) = D(x³−x−x²+1) = 3x²−2x−1

Richiami qui.

    Posso verificare il calcolo con WolframAlpha introducendo:
D((x+1)*(x-1))     D((sqrt(x)+1)*(sqrt(x)-1))     D((sqrt(x)+1)*(sqrt(x)-1)*(x^2-1))