Arcsec x'in Türevi

Arcsec x'in Türevi Nedir ?

Arcsec x'in türevi 1/(|x|.√x²-1)'dir.

$(arcsecx{)}^{\prime }=\frac{1}{|x|.\sqrt{x^2-1}}$

$\frac{d}{dx}(arcsecx)=\frac{1}{|x|.\sqrt{x^2-1}}$

Arcsec x'in Türevinin İspatı

1. Yol

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

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

$arcsec(x+h)=U$, $secU=x+h$

$arcsecx=V$, $secV=x$

$secU-secV=\cancel{x}+h-\cancel{x}$

$secU-secV=h$

$h\to 0$, $secU-secV\to 0$, $secU\to secV$, $U\to V$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }\frac{U-V}{secU-secV}$

$\boxed{seca-secb=\frac{cosb-cosa}{cosa.cosb}}$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }\frac{U-V}{\frac{cosV-cosU}{cosU.cosV}}$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }\frac{(U-V).cosU.cosV}{cosV-cosU}$

$\boxed{cosp-cosq=-2.sin\frac{p+q}{2}.sin\frac{p-q}{2}}$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }\frac{(U-V).cosU.cosV}{-2.sin\frac{V+U}{2}.sin\frac{V-U}{2}}$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }\frac{(U-V).cosU.cosV}{-2.sin\frac{U+V}{2}.sin\frac{-(U-V)}{2}}$

$\boxed{sin(-x)=-sinx}$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }\frac{(U-V).cosU.cosV}{-2.sin\frac{U+V}{2}.-sin\frac{U-V}{2}}$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }\frac{(U-V).cosU.cosV}{2.sin\frac{U+V}{2}.sin\frac{U-V}{2}}$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }\frac{\frac{1}{2}.(U-V).cosU.cosV}{sin\frac{U+V}{2}.sin\frac{U-V}{2}}$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }\frac{\frac{U-V}{2}.cosU.cosV}{sin\frac{U+V}{2}.sin\frac{U-V}{2}}$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }(\frac{\frac{U-V}{2}}{sin\frac{U-V}{2}}.\frac{cosU.cosV}{sin\frac{U+V}{2}})$

$(arcsecx{)}^{\prime }=\underset{U\to V}{\lim }\frac{\frac{U-V}{2}}{sin\frac{U-V}{2}}.\underset{U\to V}{\lim }\frac{cosU.cosV}{sin\frac{U+V}{2}}$

$U\to V$, $\frac{U-V}{2}\to 0$

$(arcsecx{)}^{\prime }=\underset{\frac{U-V}{2}\to 0}{\lim }\frac{\frac{U-V}{2}}{sin\frac{U-V}{2}}.\underset{U\to V}{\lim }\frac{cosU.cosV}{sin\frac{U+V}{2}}$

$\boxed{\underset{t\to 0}{\lim }\frac{t}{sint}=1}$

$(arcsecx{)}^{\prime }=1.\frac{cosV.cosV}{sin\frac{V+V}{2}}$

$(arcsecx{)}^{\prime }=\frac{cosV.cosV}{sin\frac{V+V}{2}}$

$(arcsecx{)}^{\prime }=\frac{co{s}^{2}V}{sin\frac{\cancel{2}V}{\cancel{2}}}$

$(arcsecx{)}^{\prime }=\frac{co{s}^{2}V}{sinV}$

$secV=x$

$sinV=\frac{\sqrt{x^2-1}}{x}$

$cosV=\frac{1}{x}$

$(arcsecx{)}^{\prime }=\frac{(\frac{1}{x}{)}^2}{\frac{\sqrt{x^2-1}}{x}}$

$(arcsecx{)}^{\prime }=\frac{\frac{1}{x^2}}{\frac{\sqrt{x^2-1}}{x}}$

$(arcsecx{)}^{\prime }=\frac{1}{\cancel{x^2}x}.\frac{\cancel{x}1}{\sqrt{x^2-1}}$

$(arcsecx{)}^{\prime }=\frac{1}{x.\sqrt{x^2-1}}$

Arcsec x fonksiyonunun herhangi bir noktasındaki teğetinin eğimi her zaman 0'dan büyük olacağı için doğal olarak herhangi bir noktasındaki türevinin değeri de 0'dan büyük olur. Bunun için yukarıdaki formüldeki paydadaki x'i mutlak değer içinde göstermemiz gerekir.

$(arcsecx{)}^{\prime }=\frac{1}{|x|.\sqrt{x^2-1}}$

2. Yol

$y=arcsecx$

$secy=x$

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

$\boxed{(secu{)}^{\prime }=u^{\prime }.secu.tanu}$

$y^{\prime }.secy.tany=1$

$y^{\prime }=\frac{1}{secy.tany}$

$secy=x$

$tany=\sqrt{x^2-1}$

$y^{\prime }=\frac{1}{x.\sqrt{x^2-1}}$

$y^{\prime }=\frac{1}{|x|.\sqrt{x^2-1}}$