derivative of e to the x

proof of the derivative of e to the x

$f(x)=e^x\Rightarrow f^{\prime }(x)=e^x$. We can prove why the derivative of $e^x$ is equal to $e^x$ (itself) in several ways, some of which we will show below.

Way 1

$\underset{n\to \infty }{\lim }(1+\frac{1}{n}{)}^n=e=\mathrm{2.71828...}$

$n\to \infty$

$\frac{1}{n}\to \frac{1}{\infty }$

$\frac{1}{n}\to 0$

If $h=\frac{1}{n}$ transformation is done;

$h\to 0$

$\underset{n\to \infty }{\lim }(1+\frac{1}{n}{)}^n=e$

$\underset{\frac{1}{n}\to 0}{\lim }(1+\frac{1}{n}{)}^n=e$

$\underset{h\to 0}{\lim }(1+h{)}^{\frac{1}{h}}=e$

$\underset{h\to 0}{\lim }\sqrt[h]{1+h}=e$

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

$(e^x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{e^{x+h}-e^x}{h}$

$(e^x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{e^x(e^h-1)}{h}$

$(e^x{)}^{\prime }=\underset{h\to 0}{\lim }{e}^x.\frac{(e^h-1)}{h}$

$(e^x{)}^{\prime }=e^x.\underset{h\to 0}{\lim }\frac{(e^h-1)}{h}$

$(e^x{)}^{\prime }=e^x.\frac{\underset{h\to 0}{\lim }(e^h-1)}{\underset{h\to 0}{\lim }h}$

$(e^x{)}^{\prime }=e^x.\frac{\underset{h\to 0}{\lim }{e}^h-\underset{h\to 0}{\lim }1}{\underset{h\to 0}{\lim }h}$

$(e^x{)}^{\prime }=e^x.\frac{(\underset{h\to 0}{\lim }\sqrt[h]{1+h}{)}^h-\underset{h\to 0}{\lim }1}{\underset{h\to 0}{\lim }h}$

$(e^x{)}^{\prime }=e^x.\frac{\underset{h\to 0}{\lim }\sqrt[h]{(1+h{)}^h}-\underset{h\to 0}{\lim }1}{\underset{h\to 0}{\lim }h}$

$(e^x{)}^{\prime }=e^x.\frac{\underset{h\to 0}{\lim }(1+h)-\underset{h\to 0}{\lim }1}{\underset{h\to 0}{\lim }h}$

$(e^x{)}^{\prime }=e^x.\frac{\underset{h\to 0}{\lim }(\cancel{1}+h-\cancel{1})}{\underset{h\to 0}{\lim }h}$

$(e^x{)}^{\prime }=e^x.\frac{\cancel{\underset{h\to 0}{\lim }h}}{\cancel{\underset{h\to 0}{\lim }h}}$

$(e^x{)}^{\prime }=e^x.1$

$(e^x{)}^{\prime }=e^x$

Way 2

$f(x)=e^x$

$lnf(x)=ln{e}^x$

$lnf(x)=x.lne$

$lnf(x)=x.1$

$lnf(x)=x$

$[lnf(x){]}^{\prime }=(x{)}^{\prime }$

$\boxed{f(x)=lng(x)\Rightarrow f^{\prime }(x)=\frac{g^{\prime }(x)}{g(x)}}$

$\frac{f^{\prime }(x)}{f(x)}=1$

$f^{\prime }(x)=f(x)$

$f^{\prime }(x)=e^x$

Way 3

By using the infinite series expansion of the function $e^x$, we can prove that the derivative of $e^x$ is equal to itself. The expansion of the function $e^x$ in question series form is as follows.

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+...$

$(e^x{)}^{\prime }=(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+...{)}^{\prime }$ (we take the derivative of both sides of the equation)

$(e^x{)}^{\prime }=(1{)}^{\prime }+(x{)}^{\prime }+(\frac{x^2}{2!}{)}^{\prime }+(\frac{x^3}{3!}{)}^{\prime }+(\frac{x^4}{4!}{)}^{\prime }+(\frac{x^5}{5!}{)}^{\prime }+...$

$(e^x{)}^{\prime }=0+1+\frac{2.x}{2!}+\frac{3.x^2}{3!}+\frac{4.x^3}{4!}+\frac{5.x^4}{5!}+...$

$(e^x{)}^{\prime }=1+\frac{\cancel{2}.x}{\cancel{2}.1!}+\frac{\cancel{3}.x^2}{\cancel{3}.2!}+\frac{\cancel{4}.x^3}{\cancel{4}.3!}+\frac{\cancel{5}.x^4}{\cancel{5}.4!}+...$

$(e^x{)}^{\prime }=1+\frac{x}{1}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$

$(e^x{)}^{\prime }=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...$

$(e^x{)}^{\prime }=e^x$