Tan x'in Türevi (Tanjantın Türevi)

Tan x'in Türevi Nedir ?

Tan x'in türevi, 1 + tan² x'dir.

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

Tan x'in Türevinin İspatı

1. Yol

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

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

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

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

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

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

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

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

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

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

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

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

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

$\boxed{tan0=0}$

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

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

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

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

2. Yol

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

3. Yol

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\boxed{\Downarrow }$

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

$Soru:$

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

$Cevap:$

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

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

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

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

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

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