The derivative of a^x (a to the power x) is equal to a^{x}lna where ln denotes the natural logarithm, that is, lna=log_{e}a. 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^{m}a^{n}.

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

= $a^x \cdot \log_e a$ [ 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_{e}a.

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

Therefore,

d/dx(a^{x})=a^{x} log_{e}a. |

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

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

d/dx(e^{x})=e^{x} log_{e}e = e^{x} as we know that log_{e}e=1.

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

**ALSO READ:**

## FAQs

**Q1: What is the derivative of a^x?**

**Answer:** The derivative of a^x is equal to a^x lna.