Double differentiation of e Power 100t+x?

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Answer 1 · 2018-06-12T04:09:11

If $f(x) = e^(100t+x)$ then $f''(x) = e^(100t+x)$
[ $t$ assumed to be a constant]

This question is somewhat ambiguous in that we do not know the nature of $t$ . Also the function could be $e^(100t) + x$ or $e^(100t+x)$

I will choose the latter and take $t$ to be a constant.

$:. f(x) = e^(100t+x)$

$= e^(100t) xx e^x$

Since $t$ is a constant $-> e^(100t)$ is also a constant.

$:. f'(x) = e^(100t) d/dx e^x$

Apply standard derivative

$f'(x) = e^(100t) e^x$

$=e^(100t+x) = f(x)$

In fact $f^n(x) = f(x)$

$:. f''(x) = e^(100t+x)$