# Derivative of e^(x-2)

by quasar987
Tags: derivative
 Sci Advisor HW Helper PF Gold P: 4,771 I am asked to find the derivative function of $f(x)=e^{x-2}$ using the definition. That is to say, I have to evaluate this limit, if it exists: $$\lim_{x\rightarrow x_0}\frac{e^{x-2} - e^{x_0-2}}{x-x_0} = \lim_{x\rightarrow x_0}\frac{e^{x} - e^{x_0}}{e^2 (x-x_0)}$$ How can this undeterminate form be simplified? Thanks. (The answer is $f'(x_0)=e^{x_0-2}$.)
 Sci Advisor P: 6,040 Factor out exp(x0) from the numerator. Then expand exp(x-x0) in a power series. The rest is obvious.
 Sci Advisor HW Helper PF Gold P: 4,771 Power serie not allowed, sorry.
 HW Helper P: 2,566 Derivative of e^(x-2) im used to defining derivatives in a slightly different way, but if you need to it should be easy to convert to your way: $$\frac{d}{dx}(e^{(x-2)}) = \lim_{h \rightarrow 0} \frac{e^{(x-2)+h} - e^{(x-2)}}{h}$$ $$= \lim_{h \rightarrow 0} e^{(x-2)} \frac{e^{h} -1}{h}$$ now noting that: $$e = \lim_{h \rightarrow 0} (1+h)^{1/h}$$ raising both sides to h (this is the only step I'm not sure about, but it gives the right answer): $$\lim_{h \rightarrow 0} e^h = \lim_{h \rightarrow 0} ((1+h)^{1/h})^h = \lim_{h \rightarrow 0} 1+h$$ so: $$\lim_{h \rightarrow 0} \frac{e^{h} -1}{h}= \lim_{h \rightarrow 0} \frac{1 + h -1}{h}= 1$$ which gives the answer.
Emeritus
PF Gold
P: 5,196
 Quote by StatusX im used to defining derivatives in a slightly different way, but if you need to it should be easy to convert to your way:
lol, nobody is defining derivatives in different ways...only the notation differs

$$h \equiv \Delta x \equiv x - x_0$$
 HW Helper P: 2,566 yea, thats all i meant.
 Sci Advisor HW Helper PF Gold P: 4,771 This looks nice Status, but isn't there a $e^{(x-2)}$ remaining?
 P: 461 Yes, which is multiplied by 1, giving the answer, as is easily verified using the chain rule.
Emeritus