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Numerade Educator

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Problem 25 Easy Difficulty

Differentiate the function.
$ j(x) = x^{2.4} + e^{2.4} $

Answer

$$j^{\prime}(x)=2.4 x^{1.4}+0=2.4 x^{1.4}$$

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Video Transcript

Hey, it's clear. So in humoring here, so we have J F X is equal to X to the 2.4 plus e to the 2.4. You derive this in terms of X, and we see that E to the 2.4 is zero. Since it's a constant and we're going to use the powerful for a differentiation where this part we get 2.4 x to the 2.4 minus one, which is equal to 2.4 x The 1.4