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.

Answer 1

f(x) given to us is e^(x)

Hence,

d/dx [f(x)^3] = 3 * [e^(x)]^2 * [e^(x)]'

= 3 * e^(2x) * e^(x)

= 3e^(3x)

which is the derivative of e^(3x and this is what we are finding in the answer.