Proof of Derivative of Natural Exponential function

The derivative of natural exponential function with respect to a variable is equal to natural exponential function. It can be derived by first principle in differential calculus. If $x$ a variable, the natural exponential function is written as $e^{\displaystyle x}$. From first principle, the differentiation of $e^{\displaystyle x}$ function with respect to $x$ can be proved as $e^{\displaystyle x}$ in mathematics.

Differentiation of function in Limit form

Write the derivative of a function in limit form as per definition of the derivative.

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

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

Take $f{(x)} = e^{\displaystyle x}$, then $f{(x+h)} = e^{\displaystyle x+h}$. For evaluating the derivative of the function $e^{\displaystyle x}$ with respect to $x$ from first principle, substitute the functions $f{(x+h)}$ and $f{(x)}$ in the derivative formula.

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

Simplify the exponential function

Look at the first term in the numerator of the exponential function. It has an exponent, formed by the sum of two literals. The term can be factored in exponential form by the product rule of exponents with same base.

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

$e^{\displaystyle x}$ is a common factor in both terms of the numerator. So, take it common from both the terms for simplifying the expression in the numerator further.

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

Now, factorize the function as follows.

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

Now, use the product rule of limits.

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

Evaluate Limits of Exponential functions

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

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

According to a standard result of limit of an exponential function, the limit of $\dfrac{e^{\displaystyle h}-1}{h}$ as $h$ tends to zero is equal to one.

$=\,\,\,$ $e^{\displaystyle x}$ $\times$ $1$

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

Therefore, it is proved that the derivative of a natural exponential function with respect to a variable is equal to natural exponential function.

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