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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.

User Alex Peck
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1 Answer

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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.

User MakotoE
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8.0k points
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