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

Kaylen Lawrence

Mathematics

17 June, 00:07

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.

+2

Answers (1)

Know the Answer?

Madyson Mckee · 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.

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.

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.