By applying the Product Rule twice, one can prove that if f, g, and h are differentiable, then (fgh)′=f′gh+fg′h+fgh′.

Now, in the above result, letting f=g=h yields ddx[f(x)]3=3[f(x)]2f′(x).

Use this last formula to differentiate y=e3x.

Question

By applying the Product Rule twice, one can prove that if f, g, and h are differentiable, then (fgh)′=f′gh+fg′h+fgh′.

Now, in the above result, letting f=g=h yields ddx[f(x)]3=3[f(x)]2f′(x).

Use this last formula to differentiate y=e3x.

anonymous · Answer

f(x) for us is e^(x)
so d/dx [f(x)^3] = 3 * [e^(x)]^2 * [e^(x)]'
= 3 * e^(2x) * e^(x) = 3e^(3x) which is in fact the derivative of e^(3x) as we wanted.