When getting a function from a instant slope, can't we just put the slope as the slope itself?

Question

I'm looking at video and I've seen that when we want to find a function from a instant slope, and I've realized that we need the function to equal the slope. Why can't we just put that slope as itself to make itself equal itself? For example, we need to find the function of an instant slope $1+x+\frac{1}{2}x^2$, and in order to get that, I can put $1+x+\frac{1}{2}x^2+\frac{1}{(3)(2)}$. But instead of putting $\frac{1}{(3)(2)}$, cant we just put $\frac{1}{2}x^2$ instead because it'll equal itself?

Answer 1

The equality sign holds only for the whole series. We have the series $e^x=\sum\limits_{k=0}^\infty \frac{x^k}{k!}$. Now we know that the derivative of e^x is the derivative itsself: $\left(e^x\right)'=e^x$. And the derivative of a sum is the sum of the derivatives. So we differentiate every single summand to obtain the derivative.

$$\begin{array}{|c|c|c|} \hline {\textrm{k-th summand}} & e^x & \left(e^x\right)' \\ \hline 0 & \color{blue}1 &0 \\ \hline 1 & \color{green}x & \color{blue}1 \\ \hline 2 & \color{red}{\frac12x^2} &\color{green}x \\ \hline 3 & \color{orange}{\frac16x^3} &\color{red}{\frac12x^2} \\ \hline 4 & \color{limegreen}{\frac1{24}x^4} &\color{orange}{\frac16x^3} \\ \hline 5 & \frac1{120}x^5 &\color{limegreen}{\frac1{24}x^4} \\ \hline \vdots &\vdots &\vdots \\ \end{array}$$

Since the series of $\left(e^x\right)'$ goes to infinity it contains the same summands as $e^x$.