Different method of evaluating ${d\over dx}(e^x)=e^x$

Question

$$y=e^x\tag1$$ $${dy\over dx}=e^x\tag2$$

We can use the series of $e^x$ to show that ${d\over dx}(e^x)=e^x$ $$e^x=1+x+{x^2\over 2!}+{x^3\over 3!}+\cdots\tag3$$

$${d\over dx}(1+x+{x^2\over 2!}+{x^3\over 3!}+\cdots)=0+1+{2x\over 2!}+{3x^2\over 3!}+\cdots=1+x+{x^2\over 2!}+\cdots=e^x\tag4$$

How else can you show that $${d\over dx}(e^x)=e^x$$ with another method?

Answer 1

There are many ways. I'll go with a more 'graphical' approach instead of a numerical. We know by the limit definition of the derivative that $m = \frac{\delta y}{\delta x}$, so we can choose any point $e^x$, $x \in \Bbb{R}$, and find the slope of the tangent line. For instance,

Let's choose a point $x = 1$:

When $x = 1$ $\rightarrow$ $y = e^1$. Similarly, when $x = 0$ $\rightarrow y = e^0 = 1$

This yields points $(1,e)$ and $(0,1)$ is If $y=mx+b$, then \begin{equation}y =\frac{1-0}{e-1} x +1\end{equation} ($b \textbf{ is always } 1$ for functions in the form $f(x) = a^x$ )

\begin{equation} y = mx+b \\ y =\frac{1-0}{e-1} x +1 \rightarrow y =\frac{x}{e-1} +1\\ y(e-1) = x + 1 \\ x = y(e-1) -1 \\ x = ey -y -1 \end{equation}

If we take the derivative with respect to $x$ of the equation above, \begin{equation} \frac{dy}{dx} = e\frac{dy}{dx} - \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{dy}{dx}(e - 1) \end{equation} Since we know \begin{equation}\frac{dy}{dx} = m\end{equation}, it's implied that \begin{equation}f'(x) = \frac{dy}{dx}\end{equation}, which satisfies the assumption that our slope

\begin{equation}m = \frac{\delta y}{\delta x}\end{equation}

You can further manipulate the equations above to prove that the slope of the line passing through both points is equal to $e^x$, no matter what two points you choose.

\begin{equation*} f(x) = e^x \\ f(1) = e^1 = e \end{equation*}

\begin{equation*} f'(1) = e^1 \\ f'(1) = e^1 = e \end{equation*}

Of course, you must provide a graph showing both coordinates. It is an intuitive demonstration of \begin{equation} lim _{h\rightarrow 0} \frac{f(x+h)-f(x)}{h)} \\ lim _{h\rightarrow 0} \frac{e^{x+h}-e^{x}}{h} \\ =\frac{e^{x+h}-e^x}{h}=e^x \\ =\frac{e^x e^h-e^x}{h}=e^x \\ =\frac{e^x(1)-e^x}{h}=e^x.\\ \lim_{h\to0}e^x=e^x \end{equation}

or $f'(x) = e^x$ for $f(x) = e^x$.

Answer 2

you can also prove it by differentiating via first derivative principle

$\forall x\in R$

$f'(x)=lim_{h\to 0}\ \dfrac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\to 0 \ }\dfrac{e^{x+h}-e^x}{h}$

$f'(x)=e^{x}\ lim_{h\to 0}\ \dfrac{e^{h}-1}{h}=e^x.\ ln\ e=e^x$

Answer 3

We know that $$ \lim_{h\to0}\frac{e^h-1}{h}=1. $$ Hence $$ \lim_{h\to0}\frac{e^{x+h}-e^x}{h}=e^x. $$

Answer 4

Let's use some implicit differentiation. We have

$$y=e^x,$$ or equivalently $$x=\ln(y),$$ differentiating through with respect to $x$ and then some rearranging gives $$1=\frac{1}{y}\frac{dy}{dx}\Rightarrow y=\frac{dy}{dx}\Rightarrow e^x=\frac{dy}{dx},$$ as required.

Answer 5

A simple method to see that such "$e$" exists is to consider

• $2^x$ for which we can see that the slope for the secant between $0$ and $1$ is $1$ and thus, by convexity, the slope of the tangent in $0$ is less than $1$

• $4^x$ for which we can see that the slope for the secant between $-\frac12$ and $0$ is also $1$ and thus, by convexity, the slope of the tangent in $0$ is greater than $1$

thus by continuity of the exponential we can conclude that exists $e\in \mathbb{R}$ with $2<e<4$ such that at $x=0$ $(e^x)'=1$, that is

$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$

which implies that the derivative of $e^x$ at $x_0$

$$\lim_{x\to x_0}\frac{e^{x}-e^{x_0}}{x-x_0}=\lim_{x\to x_0}e^{x_0}\frac{e^{x-x_0}-1}{x-x_0}=e^{x_0}$$