Tan x'in Türevi (Tanjantın Türevi)

Tan x'in Türevi Nedir ?

Tan x'in türevi, 1+tan² x'dir.

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

Tan x'in Türevinin İspatı

1. Yol

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

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

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

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

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

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

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

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

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

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

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

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

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

$\boxed{\tan 0=0}$

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

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

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

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

2. Yol

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

3. Yol

İki fonksiyonun bölümünün türevi formülünden yararlanarak da tan x'in türevinin, 1+tan² x'ye eşit olduğunu ispatlayabiliriz.

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

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

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

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

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

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

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

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

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

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

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

(Tanx)' = 1+Tan²x = 1/Cos²x = Sec²x

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

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

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

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

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

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

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

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

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

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

$\boxed{\Downarrow }$

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

$Soru:$

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

$Cevap:$

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

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

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

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

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

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