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 $a^{\displaystyle x}$ with respect to $x$ can be derived in differential calculus by first principle.

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.

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

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

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

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

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

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

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.

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

Now, factorize the function as two different functions.

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

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

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

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

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

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

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

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

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 natural logarithm of $a$.

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