Derivative of a^x by First Principle

The derivative of a^x (a to the power x) is equal to a^xlna where ln denotes the natural logarithm, that is, lna=log_ea. In this post, we will learn how to differentiate a^x using the limit definition.

The derivative formula of a^x is the following.

$\dfrac{d}{dx}(a^x)=a^x\ln a$.

The first principle or the limit definition of derivatives is given below:

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$ …(I)

Let us now find the derivative of a^x by the first principle.

Derivative of a^x using Limit Definition

Let f(x)=a^x. Then f(x+h)= a^x+h.

From the above definition (I) of derivatives, the derivative of a to the power x will be equal to

$\dfrac{d}{dx}(a^x)$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

= $\lim\limits_{h \to 0} \dfrac{a^{x+h}-a^x}{h}$

= $\lim\limits_{h \to 0} \dfrac{a^x \cdot a^h – a^x}{h}$ using the rule of indices a^m+n=a^maⁿ.

= $a^x \cdot \lim\limits_{h \to 0} \dfrac{a^h-1}{h}$

= $a^x \cdot \log_e a$ [ This is because $\lim\limits_{x \to 0}\dfrac{a^x-1}{x}=\log_e a$]

= $a^x \ln a$, provided that a>0. Here we have used lna=log_ea.

So the derivative of a^x is equal to a^xlog_ea provided that a>0. This is proved by the first principle of derivatives.

Therefore,

d/dx(ax)=ax logea.

Corollary: d/dx(e^x)=e^x.

We put a=e in the formula d/dx(a^x)=a^x log_ea. So we get that

d/dx(e^x)=e^x log_ee = e^x as we know that log_ee=1.

Thus, the derivative of e^x is equal to e^x.

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