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## Definition of d(a^x)/dx

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)?

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 Quote by 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)?
Note:

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

 Recognitions: Homework Help Science Advisor 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: $$\lim _{h \to 0}\frac{a^h - 1}{h} = f'(0)$$ You already have the equation: $$f'(x) = a^x\lim _{h \to 0}\frac{a^h - 1}{h}$$ Substitute 0 for x, and recognize that $a^0 = 1$, and you'll see why the equation for f'(0) holds.

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## Definition of d(a^x)/dx

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

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