# Differentiate #e^(ax)# using first principles?

# f'(x) = ae^(ax) #

And the clever readers who already know the answer can hopefully spot that we are almost there if we can show that

Once the limit has been established, then the result is evident giving:

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To differentiate (e^{ax}) using first principles, we use the definition of the derivative:

[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}]

For (f(x) = e^{ax}), the derivative (f'(x)) is:

[f'(x) = \lim_{h \to 0} \frac{e^{a(x+h)} - e^{ax}}{h}]

[= \lim_{h \to 0} \frac{e^{ax}e^{ah} - e^{ax}}{h}]

[= \lim_{h \to 0} \frac{e^{ax}(e^{ah} - 1)}{h}]

Using the limit definition of the exponential function (e^u):

[= \lim_{h \to 0} \frac{e^{ax}(1 + ah + \frac{(ah)^2}{2!} + ...) - e^{ax}}{h}]

[= \lim_{h \to 0} \frac{e^{ax} + ae^{ax}h + \frac{(ae^{ax}h)^2}{2!} + ... - e^{ax}}{h}]

[= \lim_{h \to 0} \frac{ae^{ax}h + \frac{(ae^{ax}h)^2}{2!} + ...}{h}]

[= \lim_{h \to 0} \left(ae^{ax} + \frac{ae^{ax}h}{2!} + ...\right)]

[= ae^{ax}]

So, the derivative of (e^{ax}) with respect to (x) using first principles is (ae^{ax}).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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