Tan u(x)'in Türevi

Tan u(x)'in Türevi Nedir ?

Tan u(x)'in türevi, u'(x).[1+tan² u(x)]'dir.

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

Tan u(x)'in Türevinin İspatı

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

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

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

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

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

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

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

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

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

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

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

$h=u(x+h)-u(x)$$(h\to 0)$

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

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

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

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

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

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

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

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

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

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

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

$Soru:$

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

$Cevap:$

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

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

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

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

$f^{\prime }(x)=(3+0).[1+{\tan }^2(3x+2)]$

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