What is the $100$ $100$ th derivative of $2 e^{x}$ $2e^x$ ?

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What is the $100$ $100$ th derivative of $2 e^{x}$ $2e^x$ ?

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Answer 1 · 2016-12-06T07:43:17

Well, you should memorize that the derivative of $e^{x}$ $e^x$ is $e^{x}$ $e^x$ . Therefore, $\frac{d^{n}}{{d x}^{n}} [e^{x}] = f^{(n)} (x) = e^{x}$ $(d^n)/(dx^n)[e^x] = f^((n))(x) = e^x$ .

The constant floats out front, so

$\frac{d^{n}}{{d x}^{n}} [2 e^{x}]$ $(d^n)/(dx^n)[2e^(x)]$

$= 2 \frac{d^{n}}{{d x}^{n}} [e^{x}]$ $= 2(d^n)/(dx^n)[e^(x)]$

$= 2 e^{x}$ $= 2e^x$

Since we just proved it in general, what does that tell you about the $100$ $100$ th derivative?

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What is the $100$ $100$ th derivative of $2 e^{x}$ $2e^x$ ?

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