Cot x'in Türevi (Kotanjantın Türevi)

Cot x'in Türevi Nedir ?

Cot x'in türevi, -(1+cot² x)'dir.

$\frac{d}{dx}(\cot x)=-(1+{\cot }^{2}x)$

Cot x'in Türevinin İspatı

1. Yol

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

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

$\boxed{\cot (p+q)=\frac{\cot p.\cot q-1}{\cot p+\cot q}}$

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

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

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

$(\cot x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cancel{\cot x.\cot h}-1-{\cot }^{2}x-\cancel{\cot x.\cot h}}{h.(\cot x+\cot h)}$

$(\cot x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-(1+{\cot }^{2}x)}{h.(\cot x+\cot h)}$

$\boxed{\cot x=\frac{1}{\tan x}}$

$(\cot x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-(1+{\cot }^{2}x)}{h.(\cot x+\frac{1}{\tan h})}$

$(\cot x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-(1+{\cot }^{2}x)}{h.\cot x+\frac{h}{\tan h}}$

$(\cot x{)}^{\prime }=\frac{-(1+{\cot }^{2}x)}{\underset{h\to 0}{\lim }(h.\cot x)+\underset{h\to 0}{\lim }\frac{h}{\tan h}}$

$(\cot x{)}^{\prime }=\frac{-(1+{\cot }^{2}x)}{\cot x.\underset{h\to 0}{\lim }h+\underset{h\to 0}{\lim }\frac{h}{\tan h}}$

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

$(\cot x{)}^{\prime }=\frac{-(1+{\cot }^{2}x)}{\cot x.0+1}$

$(\cot x{)}^{\prime }=\frac{-(1+{\cot }^{2}x)}{0+1}$

$(\cot x{)}^{\prime }=\frac{-(1+{\cot }^{2}x)}{1}$

$(\cot x{)}^{\prime }=-(1+{\cot }^{2}x)$

2. Yol

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

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

$\boxed{\cot x=\frac{\cos x}{\sin x}}$

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

$(\cot x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{\sin x.\cos (x+h)-\sin (x+h).\cos x}{\sin x.\sin (x+h)}}{h}$

$(\cot x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\sin x.\cos (x+h)-\sin (x+h).\cos x}{h.\sin x.\sin (x+h)}$

$\boxed{\sin (p-q)=\sin p.\cos q-\sin q.\cos p}$

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

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

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

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

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

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

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

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

$(\cot x{)}^{\prime }=-1.\frac{1}{\sin x.\sin (x+0)}$

$(\cot x{)}^{\prime }=-\frac{1}{\sin x.\sin (x+0)}$

$(\cot x{)}^{\prime }=-\frac{1}{\sin x.\sin x}$

$(\cot x{)}^{\prime }=-\frac{1}{{\sin }^{2}x}$

$\boxed{{\sin }^{2}x+{\cos }^{2}x=1}$

$(\cot x{)}^{\prime }=-(\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x})$

$(\cot x{)}^{\prime }=-(\frac{\cancel{{\sin }^{2}x}}{\cancel{{\sin }^{2}x}}+\frac{{\cos }^{2}x}{{\sin }^{2}x})$

$(\cot x{)}^{\prime }=-[1+(\frac{\cos x}{\sin x}{)}^2]$

$(\cot x{)}^{\prime }=-(1+{\cot }^{2}x)$

3. Yol

İki fonksiyonun bölümünün türevi formülünden yararlanarak da cot x'in türevinin, -(1+cot² x)'ye eşit olduğunu ispatlayabiliriz.

$\boxed{f(x)=\frac{u(x)}{v(x)}\Rightarrow f^{\prime }(x)=\frac{u^{\prime }(x).v(x)-v^{\prime }(x).u(x)}{[v(x){]}^2}}$

$\boxed{\cot x=\frac{\cos x}{\sin x}}$

$(\cot x{)}^{\prime }=(\frac{\cos x}{\sin x}{)}^{\prime }$

$(\cot x{)}^{\prime }=\frac{(\cos x{)}^{\prime }.\sin x-(\sin x{)}^{\prime }.\cos x}{{\sin }^{2}x}$

$\boxed{(\sin x{)}^{\prime }=\cos x}$ $\boxed{(\cos x{)}^{\prime }=-\sin x}$

$(\cot x{)}^{\prime }=\frac{-\sin x.\sin x-\cos x.\cos x}{{\sin }^{2}x}$

$(\cot x{)}^{\prime }=\frac{-{\sin }^{2}x-{\cos }^{2}x}{{\sin }^{2}x}$

$(\cot x{)}^{\prime }=\frac{-({\sin }^{2}x+{\cos }^{2}x)}{{\sin }^{2}x}$

$(\cot x{)}^{\prime }=-(\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x})$

$(\cot x{)}^{\prime }=-(\frac{\cancel{{\sin }^{2}x}}{\cancel{{\sin }^{2}x}}+\frac{{\cos }^{2}x}{{\sin }^{2}x})$

$(\cot x{)}^{\prime }=-[1+(\frac{\cos x}{\sin x}{)}^2]$

$(\cot x{)}^{\prime }=-(1+{\cot }^{2}x)$

(Cot x)' = -(1+Cot² x) = -1/Sin² x = -Csc² x

$(\cot x{)}^{\prime }=-(1+{\cot }^{2}x)$

$\boxed{\cot x=\frac{\cos x}{\sin x}}$

$(\cot x{)}^{\prime }=-[1+(\frac{\cos x}{\sin x}{)}^2]$

$(\cot x{)}^{\prime }=-(1+\frac{{\cos }^{2}x}{{\sin }^{2}x})$

$(\cot x{)}^{\prime }=-(\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x})$

$\boxed{{\sin }^{2}x+{\cos }^{2}x=1}$

$(\cot x{)}^{\prime }=-\frac{1}{{\sin }^{2}x}$

$(\cot x{)}^{\prime }=-(\frac{1}{\sin x}{)}^2$

$\boxed{\csc x=\frac{1}{\sin x}}$

$(\cot x{)}^{\prime }=-{\csc }^{2}x$

$\boxed{\Downarrow }$

$\boxed{f(x)=\cot u(x)\Rightarrow f^{\prime }(x)=-u^{\prime }(x).[1+{\cot }^{2}u(x)]}$ (formülün ispatını görmek için tıklayınız)

$Soru:$

$f(x)=\cot (2x^3+3x^2)\Rightarrow f^{\prime }(x)=?$

$Cevap:$

$f(x)=\cot (2x^3+3x^2)$

$f^{\prime }(x)=[\cot (2x^3+3x^2){]}^{\prime }$

$\boxed{f(x)=\cot u(x)\Rightarrow f^{\prime }(x)=-u^{\prime }(x).[1+{\cot }^{2}u(x)]}$

$f^{\prime }(x)=-(2x^3+3x^2{)}^{\prime }.[1+{\cot }^2(2x^3+3x^2)]$

$f^{\prime }(x)=-(\mathrm{3.2.}{x}^2+\mathrm{2.3.}x).[1+{\cot }^2(2x^3+3x^2)]$

$f^{\prime }(x)=-(6x^2+6x).[1+{\cot }^2(2x^3+3x^2)]$