Derivada de Tan u

¿ Cuál es la Derivada de Tan u ?

La derivada de tan u es u'.(1+tan² u).

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

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

Prueba de la Derivada de Tan u

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

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

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

$[tanu(x){]}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{sinu(x+h)}{cosu(x+h)}-\frac{sinu(x)}{cosu(x)}}{h}$

$[tanu(x){]}^{\prime }=\underset{h\to 0}{\lim }\frac{\frac{sinu(x+h).cosu(x)-cosu(x+h).sinu(x)}{cosu(x).cosu(x+h)}}{h}$

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

$[tanu(x){]}^{\prime }=\underset{h\to 0}{\lim }\frac{sinu(x+h).cosu(x)-cosu(x+h).sinu(x)}{h.cosu(x).cosu(x+h)}$

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

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

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

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

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

$h\to 0[u(x+h)-u(x)=h]$

$[tanu(x){]}^{\prime }=\underset{h\to 0}{\lim }\frac{u(x+h)-u(x)}{h}.\underset{h\to 0}{\lim }\frac{sinh}{h}.\underset{h\to 0}{\lim }\frac{1}{cosu(x).cosu(x+h)}$

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

$[tanu(x){]}^{\prime }=u^{\prime }(x).1.\frac{1}{cosu(x).cosu(x+0)}$

$[tanu(x){]}^{\prime }=u^{\prime }(x).1.\frac{1}{cosu(x).cosu(x)}$

$[tanu(x){]}^{\prime }=\frac{u^{\prime }(x).1.1}{co{s}^{2}u(x)}$

$[tanu(x){]}^{\prime }=\frac{u^{\prime }(x)}{co{s}^{2}u(x)}$

$u(x)=u$

$u^{\prime }(x)=u^{\prime }$

$(tanu{)}^{\prime }=\frac{u^{\prime }}{co{s}^{2}u}$

$(tanu{)}^{\prime }=u^{\prime }.\frac{1}{co{s}^{2}u}$

$(tanu{)}^{\prime }=u^{\prime }.(\frac{1}{cosu}{)}^2$

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

$(tanu{)}^{\prime }=u^{\prime }.se{c}^{2}u$

$(tanu{)}^{\prime }=u^{\prime }.\frac{1}{co{s}^{2}u}$

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

$(tanu{)}^{\prime }=u^{\prime }.\frac{co{s}^{2}u+si{n}^{2}u}{co{s}^{2}u}$

$(tanu{)}^{\prime }=u^{\prime }.(\frac{\cancel{co{s}^{2}u}}{\cancel{co{s}^{2}u}}+\frac{si{n}^{2}u}{co{s}^{2}u})$

$(tanu{)}^{\prime }=u^{\prime }.(1+\frac{si{n}^{2}u}{co{s}^{2}u})$

$(tanu{)}^{\prime }=u^{\prime }.[1+(\frac{sinu}{cosu}{)}^2]$

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

$(tanu{)}^{\prime }=u^{\prime }.(1+ta{n}^{2}u)=\frac{u^{\prime }}{co{s}^{2}u}=u^{\prime }.se{c}^{2}u$

Pregunta

$f(x)=tan(2x-7)\Rightarrow f^{\prime }(x)=?$

Respuesta

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

$[tan(2x-7){]}^{\prime }=(2x-7{)}^{\prime }.[1+ta{n}^2(2x-7)]$

$[tan(2x-7){]}^{\prime }=2.[1+ta{n}^2(2x-7)]$

$[tan(2x-7){]}^{\prime }=2.1+2.ta{n}^2(2x-7)$

$[tan(2x-7){]}^{\prime }=2+2.ta{n}^2(2x-7)$