Derivada de 1/Cos x

¿ Cuál es la Derivada de 1/Cos x ?

La derivada de 1/cos x es sec x.tan x.

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

$\frac{d}{dx}(\frac{1}{cosx})=secx.tanx$

Prueba de la Derivada de 1/Cos x

Método 1

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

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

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

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

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

$\boxed{cosp-cosq=-2.sin\frac{p-q}{2}.sin\frac{p+q}{2}}$

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

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

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

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

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

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

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

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

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

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

$(\frac{1}{cosx}{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{sin\frac{h}{2}}{\frac{h}{2}}.\frac{sin(x+\frac{h}{2})}{cosx.cos(x+h)}]$

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

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

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

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

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

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

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

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

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

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

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

Método 2

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

$(\frac{1}{cosx}{)}^{\prime }=\frac{(1{)}^{\prime }.cosx-(cosx{)}^{\prime }.1}{co{s}^{2}x}$

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

$(\frac{1}{cosx}{)}^{\prime }=\frac{0.cosx-(-sinx).1}{co{s}^{2}x}$

$(\frac{1}{cosx}{)}^{\prime }=\frac{0+sinx}{co{s}^{2}x}$

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

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

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

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

Método 3

$\frac{1}{cosx}=co{s}^{-1}x$

$(\frac{1}{cosx}{)}^{\prime }=(co{s}^{-1}x{)}^{\prime }$

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

$(\frac{1}{cosx}{)}^{\prime }=-1.co{s}^{-1-1}x.(cosx{)}^{\prime }$

$(\frac{1}{cosx}{)}^{\prime }=-1.co{s}^{-2}x.-sinx$

$(\frac{1}{cosx}{)}^{\prime }=co{s}^{-2}x.sinx$

$(\frac{1}{cosx}{)}^{\prime }=\frac{1}{co{s}^{2}x}.sinx$

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

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

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

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