# Derivative of e raised to the power of x with respect to x

## Formula

$\dfrac{d}{dx} e^{\displaystyle x} = e^{\displaystyle x}$

### Proof

Use the relation of derivative of a function in limits.

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

Take $f(x) = e^{\displaystyle x}$ and then $f(x+h) = e^{\displaystyle x+h}$. Substitute them in the formula to start deriving the derivative of the function $e^x$ with respect to $x$.

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

Use the product rule of exponents to split $e^{\displaystyle x+h}$ into two multiplying exponents.

$$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = \lim_{h \to 0} \dfrac{e^{\displaystyle x}e^{\displaystyle h} -e^{\displaystyle x}}{h}$$

In numerator, $e^x$ is a common function in both terms.

$$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = \lim_{h \to 0} \dfrac{e^{\displaystyle x} (e^{\displaystyle h} -1)}{h}$$

Use exponential series to expand the function $e^{\displaystyle h}$ in mathematics.

$e^{\displaystyle h} = 1 + \dfrac{h}{1!} + \dfrac{h^2}{2!} + \dfrac{h^3}{3!} + \cdots$

Replace the function $e^{\displaystyle h}$ by its expansion.

$$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = \lim_{h \to 0} e^{\displaystyle x} \times \dfrac{\Bigg[1 + \dfrac{h}{1!} + \dfrac{h^2}{2!} + \dfrac{h^3}{3!} + \cdots -1\Bigg]}{h}$$

$$\implies \require{cancel} \dfrac{d}{dx} \, e^{\displaystyle x} = \lim_{h \to 0} e^{\displaystyle x} \times \dfrac{\Bigg[\cancel{1} + \dfrac{h}{1!} + \dfrac{h^2}{2!} + \dfrac{h^3}{3!} + \cdots -\cancel{1}\Bigg]}{h}$$

$$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \lim_{h \to 0} \dfrac{\Bigg[\dfrac{h}{1!} + \dfrac{h^2}{2!} + \dfrac{h^3}{3!} + \cdots \Bigg]}{h}$$

$$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \lim_{h \to 0} \Bigg[\dfrac{\dfrac{h}{1!}}{h} + \dfrac{\dfrac{h^2}{2!}}{h} + \dfrac{\dfrac{h^3}{3!}}{h} + \cdots \Bigg]$$

$$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \lim_{h \to 0} \Bigg[\dfrac{h}{h \times 1!} + \dfrac{h^2}{h \times 2!} + \dfrac{h^3}{h \times 3!} + \cdots \Bigg]$$

$$\implies \require{cancel} \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \lim_{h \to 0} \Bigg[\dfrac{\cancel {h}}{\cancel{h} \times 1!} + \dfrac{\cancel{h^2}}{\cancel{h} \times 2!} + \dfrac{\cancel{h^3}}{\cancel{h} \times 3!} + \cdots \Bigg]$$

$$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \lim_{h \to 0} \Bigg[\dfrac{1}{1!} + \dfrac{h}{2!} + \dfrac{h^2}{3!} + \cdots\Bigg]$$

The value of factorial one is one.

$$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \lim_{h \to 0} \Bigg[\dfrac{1}{1} + \dfrac{h}{2!} + \dfrac{h^2}{3!} + \cdots\Bigg]$$

$$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \lim_{h \to 0} \Bigg[1 + \dfrac{h}{2!} + \dfrac{h^2}{3!} + \cdots\Bigg]$$

Now, substitute $h$ equals to zero to release the function from the limit.

$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \times \Bigg[1 + \dfrac{0}{2!} + \dfrac{0^2}{3!} + \cdots\Bigg]$

$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \times \Bigg[1 + \dfrac{0}{2!} + \dfrac{0^2}{3!} + \cdots\Bigg]$

$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \times [1 + 0 + 0 + \cdots]$

$\implies \dfrac{d}{dx} \, e^{\displaystyle x} = e^{\displaystyle x} \times 1$

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

It is proved that the derivative of $e$ raised to the power of $x$ is equal to $e$ raised to the power of the $x$.

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