# Definition of d(a^x)/dx

1. Feb 6, 2006

### G01

Hey. I'm having trouble understanding part of the definition of this derivative. Any help will be appreciated.

$$f(x)=a^x$$

Using the definition of a derivative, the derivative of the above function is:

$$f'(x) = \lim_{h \rightarrow 0}\frac{a^{x+h} + a^x}{h} =$$

$$\lim_{h \rightarrow 0} \frac{a^x(a^h - 1)}{h}$$

Since a^x does not depend on h it can be taken outside the limit:

$$f'(x) = a^x \lim_{h \rightarrow 0} \frac{a^h-1}{h}$$

Now here is where I get confused. The text tells me that:

$$\lim_{h \rightarrow 0} \frac{a^x(a^h - 1)}{h} = f'(0)$$ (1)

If that is true then $$f'(x) = f'(0)a^x$$, but I have no idea why equation 1 is the way it is? How is that limit equal to f'(0)?

Last edited: Feb 6, 2006
2. Feb 6, 2006

### JasonRox

Note:

$$\lim_{h \rightarrow 0} \frac{(a^h - 1)}{h} = \lim_{h \rightarrow 0} \frac{(a^{(0 + h)} - a^0)}{h} = f'(0)$$

3. Feb 6, 2006

### AKG

You've made some errors. In the second line, you should have a minus sign, not a plus sign in the numerator. Equation (1) should read:

$$\lim _{h \to 0}\frac{a^h - 1}{h} = f'(0)$$

$$f'(x) = a^x\lim _{h \to 0}\frac{a^h - 1}{h}$$
Substitute 0 for x, and recognize that $a^0 = 1$, and you'll see why the equation for f'(0) holds.