# Proof of Derivative of Exponential function by the Limits

The derivative of an exponential function $x$-th power of $a$ with respect to $x$ can be proved by the fundamental definition of the derivatives.

$\dfrac{d}{dx}{\big(a^{\displaystyle x}\big)}$ $\,=\,$ $a^{\displaystyle x} \times \log_{e}{a}$

Let us learn how to derivative the differentiation of the exponential function by the first principle of the derivatives.

### Derivative of a function in Limit form

To find the derivative of exponential function $a^{\displaystyle x}$ with respect to $x$, write the derivative of this function in limit form by the definition of the derivative.

$\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{f(x+\Delta x)-f(x)}{\Delta x}}$

It is simply written in the following form for our convenience.

$\implies$ $\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$

Suppose $f{(x)} \,=\, a^{\displaystyle x}$, then $f{(x+h)} \,=\, a^{\displaystyle x+h}$. Now, substitute them in the first principle of the derivative for evaluating the differentiation of the exponential function $a^{\displaystyle x}$ with respect to $x$ in the differential calculus.

$\implies$ $\dfrac{d}{dx}{\, (a^{\displaystyle x})}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle x+h}-a^{\displaystyle x}}{h}}$

### Simplify the exponential function

It is time to simplify the whole function. Look at the exponential expression in the numerator. It consists of two terms. In both terms, there is a common factor. Firstly, use product rule of exponents with same base to split the first term as a product of two exponential functions.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle x} \times a^{\displaystyle h}-a^{\displaystyle x}}{h}}$

In numerator, $a^{\displaystyle x}$ is a common factor in the both terms of the expression. So, it can be taken out common from them for simplifying this exponential function mathematically in the numerator.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle x}\big(a^{\displaystyle h}-1\big)}{h}}$

The mathematical function can be split as a product of two functions by factorization (or factorisation).

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \bigg(a^{\displaystyle x} \times \dfrac{a^{\displaystyle h}-1}{h}\bigg)}$

### Evaluate the Limit of the function

The limit of the function should have to be evaluated when the value of $h$ approaches zero. So, the exponential function $a$ raised to the $x$-th power is a constant. So, it can be taken out from the limit operation as per the constant multiple rule of limits.

$=\,\,\,$ $a^{\displaystyle x}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{a^{\displaystyle h}-1}{h}}$

According to the limit rule of exponential function, the limit of $a$ raised to the power of $h$ minus $1$ by $h$ as the value $h$ is closer to zero is equal to natural logarithm of $a$.

$=\,\,\,$ $a^{\displaystyle x}$ $\times$ $\log_{e}{a}$

$\therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\,\big(a^{\displaystyle x}\big)} \,=\, a^{\displaystyle x}\log_{e}{a}$

Therefore, it is proved that the derivative of exponential function $a^{\displaystyle x}$ with respect to $x$ is equal to product of the exponential function $a^{\displaystyle x}$ and the natural logarithm of $a$.

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