Derivada de Csc x

¿ Cuál es la Derivada de Csc x ?

La derivada de csc x es -csc x.cot x.

$(cscx{)}^{\prime }=-cscx.cotx$

$\frac{d}{dx}(cscx)=-cscx.cotx$

Prueba de la Derivada de Csc x

Método 1

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

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

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

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

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

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

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

$\boxed{sinp-sinq=2.sin\frac{p-q}{2}.cos\frac{p+q}{2}}$

$(cscx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.sin\frac{x-(x+h)}{2}.cos\frac{x+x+h}{2}}{h.sinx.sin(x+h)}$

$(cscx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.sin\frac{\cancel{x}-\cancel{x}-h}{2}.cos\frac{2x+h}{2}}{h.sinx.sin(x+h)}$

$(cscx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.sin\frac{-h}{2}.cos\frac{2x+h}{2}}{h.sinx.sin(x+h)}$

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

$(cscx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.-sin\frac{h}{2}.cos\frac{2x+h}{2}}{h.sinx.sin(x+h)}$

$(cscx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.-sin\frac{h}{2}.cos\frac{\cancel{2}.(x+\frac{h}{2})}{\cancel{2}}}{h.sinx.sin(x+h)}$

$(cscx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.-sin\frac{h}{2}.cos(x+\frac{h}{2})}{h.sinx.sin(x+h)}$

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

$(cscx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-sin\frac{h}{2}.cos(x+\frac{h}{2})}{\frac{h}{2}.sinx.sin(x+h)}$

$(cscx{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{-sin\frac{h}{2}}{\frac{h}{2}}.\frac{cos(x+\frac{h}{2})}{sinx.sin(x+h)}]$

$(cscx{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-sin\frac{h}{2}}{\frac{h}{2}}.\underset{h\to 0}{\lim }\frac{cos(x+\frac{h}{2})}{sinx.sin(x+h)}$

$h\to 0(\frac{h}{2}=h)$

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

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

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

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

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

$(cscx{)}^{\prime }=-\frac{cosx}{sinx.sinx}$

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

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

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

$(cscx{)}^{\prime }=-cscx.cotx$

Método 2

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

$(cscx{)}^{\prime }=(\frac{1}{sinx}{)}^{\prime }$

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

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

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

$(cscx{)}^{\prime }=\frac{0.sinx-cosx.1}{si{n}^{2}x}$

$(cscx{)}^{\prime }=\frac{0-cosx}{si{n}^{2}x}$

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

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

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

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

$(cscx{)}^{\prime }=-cscx.cotx$

Método 3

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

$cscx=si{n}^{-1}x$

$(cscx{)}^{\prime }=(si{n}^{-1}x{)}^{\prime }$

$\boxed{(u^n{)}^{\prime }=n.u^{n-1}.u^{\prime }}$

$(cscx{)}^{\prime }=-1.si{n}^{-1-1}x.(sinx{)}^{\prime }$

$(cscx{)}^{\prime }=-1.si{n}^{-2}x.cosx$

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

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

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

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

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

$(cscx{)}^{\prime }=-cscx.cotx$