# How to find instantaneous rate of change for

How to find instantaneous rate of change for $f\left(x\right)={e}^{x}$ at x=1?
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grubijanebb
The instantaneous rate of change is also known as the derivative. It is analogous to the slope of the tangent line at a point, as well.
We might say that the instantaneous rate of change of f at x=a is equal to f'(a).
Here, we have to know that the derivative of ${e}^{x}$ is itself it's a unique and very important function, especially in calculus. That is:
$f\left(x\right)={e}^{x}⇒{f}^{\prime }\left(x\right)={e}^{x}$
So, the instantaneous rate of change of f at x=1 is f'(1), and we see that:
${f}^{\prime }\left(x\right)={e}^{x}⇒{f}^{\prime }\left(1\right)={e}^{1}=e$
The instantaneous rate of change of f at x=1 is e, which is a transcendental number approximately equal to 2.7
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Aganippe76
Given:
$f\left(x\right)={e}^{x}$ at x=1
$\frac{d}{dx}\left({e}^{x}\right)={e}^{x}$
x=1
$\left(\frac{d}{dx}\left({e}^{x}\right)\right){|}_{\left(x=1\right)}=\left({e}^{x}\right){|}_{\left(x=1\right)}=e$
$f\left(x\right)={e}^{x}$ at x=1 is e