Definition of d(a^x)/dx

G01

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

[tex] f(x)=a^x [/tex]

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

[tex]f'(x) = \lim_{h \rightarrow 0}\frac{a^{x+h} + a^x}{h} = [/tex]

[tex] \lim_{h \rightarrow 0} \frac{a^x(a^h - 1)}{h} [/tex]

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

[tex] f'(x) = a^x \lim_{h \rightarrow 0} \frac{a^h-1}{h} [/tex]

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

[tex] \lim_{h \rightarrow 0} \frac{a^x(a^h - 1)}{h} = f'(0) [/tex] (1)

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

JasonRox

Note:

[tex] \lim_{h \rightarrow 0} \frac{(a^h - 1)}{h} = \lim_{h \rightarrow 0} \frac{(a^{(0 + h)} - a^0)}{h} = f'(0)[/tex]

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:

[tex]\lim _{h \to 0}\frac{a^h - 1}{h} = f'(0)[/tex]

You already have the equation:

[tex]f'(x) = a^x\lim _{h \to 0}\frac{a^h - 1}{h}[/tex]

Substitute 0 for x, and recognize that [itex]a^0 = 1[/itex], and you'll see why the equation for f'(0) holds.

G01

Ahhh icic that was simpler than i thought. Thank you.