Cot x'in Türevi

Cot x'in Türevi Nedir ?

Cot x'in türevi -(1+cot² x)'dir.

$(cotx{)}^{\prime }=-(1+co{t}^{2}x)=-cs{c}^{2}x=-\frac{1}{si{n}^{2}x}$

$\frac{d}{dx}(cotx)=-(1+co{t}^{2}x)=-cs{c}^{2}x=-\frac{1}{si{n}^{2}x}$

Cot x'in Türevinin İspatı

1. Yol

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

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

$\boxed{cot(p+q)=\frac{cotp.cotq-1}{cotp+cotq}}$

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

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

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

$(cotx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{-(1+co{t}^{2}x)}{cotx+coth}}{h}$

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

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

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

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

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

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

$\boxed{\underset{t\to 0}{\lim }(t.cott)=1}$

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

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

$(cotx{)}^{\prime }=\frac{-(1+co{t}^{2}x)}{1}$

$(cotx{)}^{\prime }=-(1+co{t}^{2}x)$

2. Yol

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

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

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

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

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

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

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

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

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

$\boxed{sin(-x)=-sinx}$

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

$(cotx{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{1}{h}.\frac{-sinh}{sinx.sin(x+h)}]$

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

$(cotx{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{-sinh}{h}.\frac{1}{sinx.sin(x+h)}]$

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

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

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

$(cotx{)}^{\prime }=-1.\frac{1}{sinx.sin(x+0)}$

$(cotx{)}^{\prime }=-\frac{1}{sinx.sin(x+0)}$

$(cotx{)}^{\prime }=-\frac{1}{sinx.sinx}$

$(cotx{)}^{\prime }=-\frac{1}{si{n}^{2}x}$

3. Yol

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

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

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

$(cotx{)}^{\prime }=\frac{(cosx{)}^{\prime }.\sin x-(sinx{)}^{\prime }.cosx}{si{n}^{2}x}$

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

$(cotx{)}^{\prime }=\frac{-sinx.sinx-cosx.cosx}{si{n}^{2}x}$

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

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

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

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

$(cotx{)}^{\prime }=-\frac{1}{si{n}^{2}x}$

$(cotx{)}^{\prime }=-(\frac{1}{sinx}{)}^2$

$\boxed{\frac{1}{sinx}=cscx}$

$(cotx{)}^{\prime }=-cs{c}^{2}x$

Soru

$y=cot5x\Rightarrow y^{\prime }=?$

1. Cevap

$y=cot5x$

$dy=d(cot5x)$

$dy=(cot5x{)}^{\prime }dx$

$\frac{dy}{dx}=(cot5x{)}^{\prime }$

$u=5x$

$du=d(5x)$

$du=(5x{)}^{\prime }dx$

$du=5dx$

$\frac{du}{dx}=5$

$y=cotu$

$dy=d(cotu)$

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

$dy=-(1+co{t}^{2}u)du$

$\frac{dy}{du}=-(1+co{t}^{2}u)$

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

$5.-(1+co{t}^{2}u)=(cot5x{)}^{\prime }$

$-5.(1+co{t}^{2}u)=(cot5x{)}^{\prime }$

$-5.(1+co{t}^{2}5x)=(cot5x{)}^{\prime }$

2. Cevap

$(cotu{)}^{\prime }=-u^{\prime }.(1+co{t}^{2}u)$

$(cot5x{)}^{\prime }=-(5x{)}^{\prime }.(1+co{t}^{2}5x)$

$(cot5x{)}^{\prime }=-5.(1+co{t}^{2}5x)$