Derivative of a^x using First Principle of Derivatives

The Mathematician — Thu, 24 Nov 2022 13:00:46 +0000

Using the First Principle of Derivatives, we will prove that the derivative of

a^x

is equal to

a^x\ln(a)

.

Proof. Let

f(x) = a^x

. Note that

a^x = e^{x\ln(a)}

. Then

\begin{align*} f'(x) &= \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \rightarrow 0} \frac{a^{x + h} - a^x}{h} \\ &= \lim_{h \rightarrow 0} \frac{e^{(x+h)\ln(a)} - e^{x\ln(a)}}{h} \\ &= \lim_{h \rightarrow 0} \frac{e^{x\ln(a)}(e^{h\ln(a)} - 1)}{h} \\ &= e^{x\ln(a)} \cdot \lim_{h \rightarrow 0} \frac{e^{h\ln(a)} - 1}{h} \\ &= a^x \cdot \lim_{h \rightarrow 0} \frac{e^{h\ln(a)} - 1}{h}. \end{align*}

Now we need to prove what the next limit is:

\begin{align*} \lim_{h \rightarrow 0} \frac{e^{h\ln(a)} - 1}{h}, \end{align*}

We know that the Maclaurin series of

e^x

\begin{align*} e^x &= \sum_{n = 0}^{\infty} \frac{x^n}{n!} \\ &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \end{align*}

In our case, this means we have

x = h\ln(a)

, so we have:

\begin{align*} e^{h\ln(a)} &= \sum_{n = 0}^{\infty} \frac{(h\ln(a))^n}{n!} \\ &= 1 + h\ln(a) + \frac{(h\ln(a))^2}{2!} + \frac{(h\ln(a))^3}{3!} + \cdots \end{align*}

Further, we will subtract the Maclaurin series with 1:

\begin{align*} e^{h\ln(a)} - 1 &= \sum_{n = 1}^{\infty} \frac{(h\ln(a))^n}{n!} \\ &= h\ln(a) + \frac{(h\ln(a))^2}{2!} + \frac{(h\ln(a))^3}{3!} + \cdots. \end{align*}

Now we divide it out with

h

\begin{align*} \frac{e^{h\ln(a)} - 1}{h} &= \sum_{n = 1}^{\infty} \frac{h^{n-1}\ln(a)^n}{n!} \\ &= \ln(a) + \frac{h\ln(a)^2}{2!} + \frac{h^2\ln(a)^3}{3!} + \cdots \end{align*}

So we get:

\begin{align*} \lim_{h \rightarrow 0} \frac{e^{h\ln(a)} - 1}{h} &= \lim_{h \rightarrow 0} \sum_{n = 1}^{\infty} \frac{h^{n-1}\ln(a)^n}{n!} \\ &= \lim_{x \rightarrow 0} \bigg(\ln(a) + \frac{h\ln(a)^2}{2!} + \frac{h^2\ln(a)^3}{3!} + \cdots\bigg) \\ &= \ln(a) + \lim_{x \rightarrow 0} \bigg(\frac{h\ln(a)^2}{2!} + \frac{h^2\ln(a)^3}{3!} + \cdots\bigg) \\ &= \ln(a) + 0 + 0 + 0 + \cdots \\ &= \ln(a). \end{align*}

Substituting everything, we get:

\begin{align*} f'(x) = a^x \cdot \lim_{h \rightarrow 0} \frac{e^{h\ln(a)} - 1}{h} = a^x \ln(a). \end{align*}

So, the derivative of

a^x

a^x\ln(a)

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