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(a) Use the Product Rule twice to prove that it $ f,g, $ and $ h $ are differentiable. then $ (fgh)' = f'gh + fgh'. + fgh'. $

(b) Taking $ f = g = h $ in part (a), show that

$ \frac {d}{dx}[f(x)]^3=3[f(x)]^2f'(x) $

(c) Use part (b) to differentiate $ y = e^{3x}. $

(a) $=f^{\prime} g h+f g^{\prime} h+f g h^{\prime}$

(b) $=3[f(x)]^{2} f^{\prime}(x)$

(c) $=3 e^{3 x}$

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