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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 :

syed514

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.

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