Tan x'in Türevi

Tan x'in Türevi Nedir ?

Tan x'in türevi 1+tan² x'dir.

$(tanx{)}^{\prime }=1+ta{n}^{2}x=se{c}^{2}x=\frac{1}{co{s}^{2}x}$

$\frac{d}{dx}(tanx)=1+ta{n}^{2}x=se{c}^{2}x=\frac{1}{co{s}^{2}x}$

Tan x'in Türevinin İspatı

1. Yol

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

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

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

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

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

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

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

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

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

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

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

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

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

$(tanx{)}^{\prime }=1.\frac{1+ta{n}^{2}x}{1-tanx.tan0}$

$(tanx{)}^{\prime }=\frac{1+ta{n}^{2}x}{1-tanx.tan0}$

$\boxed{tan0=0}$

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

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

$(tanx{)}^{\prime }=\frac{1+ta{n}^{2}x}{1}$

$(tanx{)}^{\prime }=1+ta{n}^{2}x$

2. Yol

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

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

$\boxed{tanx=\frac{sinx}{cosx}}$

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

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

$\boxed{sinp.cosq-cosp.sinq=sin(p-q)}$

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

$(tanx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{sinh}{cosx.cos(x+h)}}{h}$

$(tanx{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{1}{h}.\frac{sinh}{cosx.cos(x+h)}]$

$(tanx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{sinh}{h.cosx.cos(x+h)}$

$(tanx{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{sinh}{h}.\frac{1}{cosx.cos(x+h)}]$

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

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

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

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

$(tanx{)}^{\prime }=\frac{1}{cosx.cosx}$

$(tanx{)}^{\prime }=\frac{1}{co{s}^{2}x}$

3. Yol

$tanx=\frac{sinx}{cosx}$

$(tanx{)}^{\prime }=(\frac{sinx}{cosx}{)}^{\prime }$

$\boxed{(\frac{u}{v}{)}^{\prime }=\frac{u^{\prime }.v-v^{\prime }.u}{v^2}}$

$(tanx{)}^{\prime }=\frac{(sinx{)}^{\prime }.cosx-(cosx{)}^{\prime }.sinx}{co{s}^{2}x}$

$\boxed{(sinx{)}^{\prime }=cosx}$ $\boxed{(cosx{)}^{\prime }=-sinx}$

$(tanx{)}^{\prime }=\frac{cosx.cosx-(-sinx).sinx}{co{s}^{2}x}$

$(tanx{)}^{\prime }=\frac{co{s}^{2}x+si{n}^{2}x}{co{s}^{2}x}$

$\boxed{co{s}^{2}x+si{n}^{2}x=1}$

$(tanx{)}^{\prime }=\frac{1}{co{s}^{2}x}$

$(tanx{)}^{\prime }=(\frac{1}{cosx}{)}^2$

$\boxed{\frac{1}{cosx}=secx}$

$(tanx{)}^{\prime }=se{c}^{2}x$

Soru

$y=tan(3-4x)\Rightarrow y^{\prime }=?$

1. Cevap

$y=tan(3-4x)$

$dy=d[tan(3-4x)]$

$dy=[tan(3-4x){]}^{\prime }dx$

$\frac{dy}{dx}=[tan(3-4x){]}^{\prime }$

$u=3-4x$

$du=d(3-4x)$

$du=(3-4x{)}^{\prime }dx$

$du=-4dx$

$\frac{du}{dx}=-4$

$y=tanu$

$dy=d(tanu)$

$dy=(tanu{)}^{\prime }du$

$dy=(1+ta{n}^{2}u)du$

$\frac{dy}{du}=1+ta{n}^{2}u$

$\frac{\cancel{du}}{dx}.\frac{dy}{\cancel{du}}=\frac{dy}{dx}$

$-4.(1+ta{n}^{2}u)=[tan(3-4x){]}^{\prime }$

$-4.[1+ta{n}^2(3-4x)]=[tan(3-4x){]}^{\prime }$

2. Cevap

$(tanu{)}^{\prime }=u^{\prime }.(1+ta{n}^{2}u)$

$[tan(3-4x){]}^{\prime }=(3-4x{)}^{\prime }.[1+ta{n}^2(3-4x)]$

$[tan(3-4x){]}^{\prime }=-4.[1+ta{n}^2(3-4x)]$