# Proof of Derivative of Exponential function

The derivative of exponential function with respect to a variable is equal to the product of the exponential function and natural logarithm of base of the exponential function. The differentiation of exponential function $a^x$ with respect to $x$ can be derived in differential calculus by first principle.

### Differentiation of function in Limit form

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

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

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

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

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

### Simplify the exponential function

Use product rule of exponents with same base to split the first term in the numerator of the exponential function.

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

In numerator, $a^x$ 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.

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

Now, factorize the function as two different functions.

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

Use, the product rule of limits for calculating the limit of the function by the product of their limits.

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

### Evaluate Limits of Exponential functions

Now, evaluate the limit of the exponential function $a^x$ as $h$ approaches zero by using direct substitution method.

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

As per the limit of an exponential function, the limit of $\dfrac{a^h}-1}{h$ as $h$ approaches zero is equal to natural logarithm of $a$.

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

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

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

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