Tan 2x'in Türevi

Tan 2x'in Türevi Nedir ?

Tan 2x'in türevi, 2.(1+tan² 2x)'dir.

$\frac{d}{dx}(\tan 2x)=2.(1+{\tan }^{2}2x)$

Tan 2x'in Türevinin İspatı

1. Yol

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$h=2h$$(h\to 0)$

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

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

$(\tan 2x{)}^{\prime }=\mathrm{2.1.}\frac{1+{\tan }^{2}2x}{1-\tan 2x.\tan 0}$

$\boxed{\tan 0=0}$

$(\tan 2x{)}^{\prime }=\mathrm{2.1.}\frac{1+{\tan }^{2}2x}{1-\tan 2x.0}$

$(\tan 2x{)}^{\prime }=\mathrm{2.1.}\frac{1+{\tan }^{2}2x}{1-0}$

$(\tan 2x{)}^{\prime }=\mathrm{2.1.}\frac{1+{\tan }^{2}2x}{1}$

$(\tan 2x{)}^{\prime }=\mathrm{2.1.}(1+{\tan }^{2}2x)$

$(\tan 2x{)}^{\prime }=2.(1+{\tan }^{2}2x)$

2. Yol

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

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

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

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

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

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

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

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

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

$(\tan 2x{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{\sin 2h}{h.\cos 2x.\cos (2x+2h)}.\frac{2}{2}]$

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

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

$(\tan 2x{)}^{\prime }=2.\underset{h\to 0}{\lim }[\frac{\sin 2h}{2h}.\frac{1}{\cos 2x.\cos (2x+2h)}]$

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

$h=2h$$(h\to 0)$

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

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

$(\tan 2x{)}^{\prime }=\mathrm{2.1.}\frac{1}{\cos 2x.\cos (2x+0)}$

$(\tan 2x{)}^{\prime }=\mathrm{2.1.}\frac{1}{\cos 2x.\cos 2x}$

$(\tan 2x{)}^{\prime }=\mathrm{2.1.}\frac{1}{{\cos }^{2}2x}$

$(\tan 2x{)}^{\prime }=2.\frac{1}{{\cos }^{2}2x}$

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

$(\tan 2x{)}^{\prime }=2.\frac{{\cos }^{2}2x+{\sin }^{2}2x}{{\cos }^{2}2x}$

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

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

$(\tan 2x{)}^{\prime }=2.(1+{\tan }^{2}2x)$

(Tan 2x)' = 2.(1+Tan² 2x) = 2/Cos² 2x = 2.Sec² 2x

$(\tan 2x{)}^{\prime }=2.(1+{\tan }^{2}2x)$

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

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

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

$(\tan 2x{)}^{\prime }=2.(\frac{{\cos }^{2}2x+{\sin }^{2}2x}{{\cos }^{2}2x})$

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

$(\tan 2x{)}^{\prime }=2.\frac{1}{{\cos }^{2}2x}$

$(\tan 2x{)}^{\prime }=2.(\frac{1}{\cos 2x}{)}^2$

$\boxed{\sec x=\frac{1}{\cos x}}$

$(\tan 2x{)}^{\prime }=2.{\sec }^{2}2x$