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Derivative of the exponential function from the limit definition

Recall that the derivative of a function ff is defined by the limit f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} whenever this limit exists.

Let f(x)=exf(x) = e^x. Using the limit definition of the derivative, f′(x)=lim⁡h→0f(x+h)−f(x)h,f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, compute f′(x)f'(x). You may use the fact that lim⁡h→0eh−1h=1\displaystyle \lim_{h \to 0} \frac{e^h - 1}{h} = 1.

Hint

Write f(x+h)=ex+h=ex⋅ehf(x+h) = e^{x+h} = e^x \cdot e^h using exponent rules, then factor exe^x out of the numerator so that the difference quotient becomes ex⋅eh−1he^x \cdot \frac{e^h - 1}{h}.

Step-by-step solution

Start with the difference quotient: f(x+h)−f(x)h=ex+h−exh.\frac{f(x+h)-f(x)}{h} = \frac{e^{x+h} - e^x}{h}. Since ex+h=exehe^{x+h} = e^x e^h, factor out exe^x from the numerator to get exeh−exh=ex⋅eh−1h.\frac{e^x e^h - e^x}{h} = e^x \cdot \frac{e^h - 1}{h}. Note that exe^x does not depend on hh, so it can be pulled outside the limit.

Now take the limit as h→0h \to 0: f′(x)=lim⁡h→0ex⋅eh−1h=exlim⁡h→0eh−1h.f'(x) = \lim_{h\to 0} e^x \cdot \frac{e^h-1}{h} = e^x \lim_{h\to 0}\frac{e^h-1}{h}. Using the given fact that lim⁡h→0eh−1h=1\lim_{h\to 0} \frac{e^h-1}{h} = 1, this simplifies to f′(x)=ex⋅1=exf'(x) = e^x \cdot 1 = e^x.

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