Tag Archives: differentiating e

February 25, 2025 · 9:28 am

Differentiating f(x)=e^x

We are going to differentiate f(x)=e^x from first principals.

Remember the definition of a derivative is

(1)   \begin{equation*}f'(x)=\lim_{\limits{h\to 0}}\frac{f(x+h)-f(x)}{h}\end{equation*}

If f(x)=e^x, then

    \begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^{x+h}-e^x}{h}\end{equation}

    \begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x\times e^h-e^x}{h}\end{equation}

    \begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x(e^h-1)}{h}\end{equation}

    \begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}

Let’s think about \lim_{\limits{h \to 0}}\frac{e^h-1}{h}

Remember e is defined as \lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n

(2)   \begin{equation*}\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation*}

Let y=e^h-1, as h \to 0, e^h-1 \to 1-1=0 hence y \to 0

If y=e^h-1, then h=ln(y+1)

(3)   \begin{equation*}\lim_{\limits{y \to 0}}\frac{y}{ln(y+1)}\end{equation*}

We are going to rewrite the equation 3 as

(4)   \begin{equation*}\lim_{\limits{y \to 0}}\frac{\frac{1}{ln(y+1)}}{y}\end{equation*}

And then we can write equation 4 as

(5)   \begin{equation*}{\lim_{\limits{y \to 0}}{\frac{1}{\frac{1}{y}ln(y+1)}\end{equation*}

Using log laws we can write equation 5 as

(6)   \begin{equation*}\lim_{\limits{y \to 0}}\frac{1}{ln(y+1)^{\frac{1}{y}}}\end{equation*}

Let y=\frac{1}{n}

As y \to 0, n \to \infty

equation 6 becomes

(7)   \begin{equation*}\lim_{\limits{n \to \infty}}\frac{1}{ln(\frac{1}{n}+1)^n}\end{equation*}

We can move the limit to inside the natural log

(8)   \begin{equation*}\frac{1}{ln(\lim_{\limits{n \to \infty}}(\frac{1}{n}+1)^n)}\end{equation*}

And we know from the definition of e that e=\lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n

Hence, equation 8 is

(9)   \begin{equation*}\frac{1}{ln(e)}=1\end{equation*}

Back to our derivative

    \begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}

We know that \lim_{\limits{h \to 0}}{\frac{e^h-1}{h}=1 hence

    \begin{equation*}f'(x)=e^x\end{equation}

Leave a Comment

Filed under Calculus, Differentiation, Year 12 Mathematical Methods

Tagged as