e üzeri ax'in türevi

e üzeri ax'in türevi nedir ?

$f(x)=e^{ax}\Rightarrow f^{\prime }(x)=a.e^{ax}$ olur.

e üzeri ax'in türevinin ispatı

$e^{ax}$'in türevinin neden $a.e^{ax}$'e eşit olduğunu aşağıda bazılarını göstereceğimiz bir kaç yoldan ispatlayabiliriz.

1. Yol

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

$n\to \infty$

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

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

$h=\frac{1}{n}$ dönüşümü yapılırsa;

$h\to 0$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\boxed{x^n-1=(x-1).(x^{n-1}+x^{n-2}+x^{n-3}+...+x^3+x^2+x+1)}$

$(e^{ax}{)}^{\prime }=e^{ax}.\frac{\underset{h\to 0}{\lim }(\cancel{1}+h-\cancel{1}).[(1+h{)}^{a-1}+(1+h{)}^{a-2}+(1+h{)}^{a-3}+...+1]}{\underset{h\to 0}{\lim }h}$

$(e^{ax}{)}^{\prime }=e^{ax}.\underset{h\to 0}{\lim }\frac{\cancel{h}.[(1+h{)}^{a-1}+(1+h{)}^{a-2}+(1+h{)}^{a-3}+...+1)]}{\cancel{h}}$

$(e^{ax}{)}^{\prime }=e^{ax}.[(1+0{)}^{a-1}+(1+0{)}^{a-2}+(1+0{)}^{a-3}+...+1]$

$(e^{ax}{)}^{\prime }=e^{ax}.(1^{a-1}+1^{a-2}+1^{a-3}+...+1)$

$(e^{ax}{)}^{\prime }=e^{ax}.\underset{\text{a tane}}{\underbrace{(1+1+1+...+1)}}$

$(e^{ax}{)}^{\prime }=a.e^{ax}$

2. Yol

$f(x)=e^{ax}$

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

$lnf(x)=ax.lne$

$lnf(x)=ax.1$

$lnf(x)=ax$

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

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

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

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

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

3. Yol

$e^{ax}$ fonksiyonunun sonsuz seri şeklindeki açılımından faydalanarak da $e^{ax}$'in türevinin $a.e^{ax}$ 'e eşit olduğunu ispatlayabiliriz. $e^{ax}$ fonksiyonunun soru seri şeklindeki açılımı aşağıdaki gibidir.

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

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

$(e^{ax}{)}^{\prime }=(1+a.x+\frac{a^2.x^2}{2!}+\frac{a^3.x^3}{3!}+\frac{a^4.x^4}{4!}+\frac{a^5.x^5}{5!}+...{)}^{\prime }$ (eşitliğin her iki tarafının da türevini alırız)

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

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

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

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

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

$(e^{ax}{)}^{\prime }=a.e^{ax}$