Why does the formula to get $(e^{x})'s$ slope differ from itself?

Question

right now, and I've noticed that when the professor shows the formula to get the slope of $e^x$, it's different from the formula to get $e^x$. Why does this happen when the slope of $e^x$ is $e^x$? There's no reason it should be different. The formula to get $e^x$ is shown as $\frac{x^n}{(n)(n-1)\cdots}$ but the formula to get $dy/dx$ of $e^x$ is $\frac{x^{n-1}}{(n-1)\cdots(1)}+\frac{x^n}{n!}$.

Answer 1

They are, in fact, the same.

By definition, we have $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac12x^2+\cdots+\underbrace{\frac{x^{n-1}}{(n-1)!}}_{(n-1)\text{th term}}+\underbrace{\frac{x^n}{n!}}_{n\text{th term}}+\underbrace{\frac{x^{n+1}}{(n+1)!}}_{(n+1)\text{th term}}+\cdots$$

Under the linearity of differentiation, the derivative of a sum is the sum of the derivatives of each term. By differentiating the $n$th term in the series, we get the $(n-1)$th term of the original series. For example, differentiating the $n$th term in the above series using the power rule, we get $$\frac{d}{dx}\left(\frac{x^n}{n!}\right)=\frac{nx^{n-1}}{n!}=\frac{x^{n-1}}{(n-1)!}$$ which we can see is the original $(n-1)$th term. Differentiating the $(n+1)$th term, we get $$\frac{d}{dx}\left(\frac{x^{n+1}}{(n+1)!}\right)=\frac{(n+1)x^n}{(n+1)!}=\frac{x^n}{n!}$$ which we can see is the original $n$th term.

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As an aside, in case you had trouble seeing why the rational expression cancels out so nicely, recall the definition of the factorial $$n!=n(n-1)(n-2)\cdots(3)(2)(1)=n(n-1)!.$$

As such, the last step in the above section looks something like this: $$\require{cancel}\frac{(n+1)x^n}{(n+1)!}=\frac{\cancel{(n+1)}x^n}{\cancel{(n+1)}(n!)}=\frac{x^n}{n!}$$