The derivative of e^x is e^x.
To show this, we use the definition of the derivative:
lim(h->0) [(e^(x+h) – e^x) / h]
= lim(h->0) [e^x(e^h – 1) / h]
Let u = e^h – 1, then as h -> 0, u -> 0
So, we can rewrite the above limit as:
lim(u->0) [e^x(u) / ln(1+u)]
Now, we can use L’Hopital’s rule to evaluate the limit:
lim(u->0) [e^x(u) / ln(1+u)] = lim(u->0) [e^x * 1 / (1+u)] = e^x
Therefore, d/dx[e^x] = e^x.
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