Ejercicio resuelto de Matemáticas I

Aplicar la regla de la cadena para calcular la derivada de las siguientes funciones:

a) $f(x)=\mathrm{arctg}\mathrm{sen}\cos \mathrm{tg}x$

b) $f(x)=\ln \sqrt{\mathrm{sen}{e}^{x^2}}$

c) $f(x)=\mathrm{arctg}\mathrm{sen}{e}^{\sqrt{x}}$

d) $f(x)={\mathrm{sen}}^2{\cos }^3(4x^3+3{)}^2$

e) $f(x)=2\cdot {\mathrm{tg}}^2{\ln }^3(x^4+7{)}^5$

f) $f(x)={\ln }^3\sqrt[3]{{\mathrm{arctg}}^4{e}^{\frac{x^2+\ln x}{x^2}}}$

Estos casos son composiciones de varias funciones; aplicamos repetidamente la regla de la cadena (derivamos de fuera hacia dentro).

$\mathbf{\text{a)}}$ $f(x)=\mathrm{arctg}(\mathrm{sen}(\cos (\mathrm{tg}x)))$. Derivando capa a capa:

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{1}{1+{\mathrm{sen}}^2(\cos (\mathrm{tg}x))}}\cdot \cos (\cos (\mathrm{tg}x))\cdot (-\mathrm{sen}(\mathrm{tg}x))\cdot {\displaystyle \frac{1}{{\cos }^{2}x}}$

Es decir, $f^{\mathrm{\prime }}(x)={\displaystyle \frac{-\cos (\cos (\mathrm{tg}x))\mathrm{sen}(\mathrm{tg}x)}{[1+{\mathrm{sen}}^2(\cos (\mathrm{tg}x))]{\cos }^{2}x}}$.

$\mathbf{\text{b)}}$ $f(x)=\ln \sqrt{\mathrm{sen}{e}^{x^2}}={\displaystyle \frac{1}{2}}\ln (\mathrm{sen}{e}^{x^2})$.

Con $g=\mathrm{sen}{e}^{x^2}$ se tiene $g^{\mathrm{\prime }}=\cos e^{x^2}\cdot e^{x^2}\cdot 2x$, así que:

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{2x{e}^{x^2}\cos e^{x^2}}{\mathrm{sen}{e}^{x^2}}}=x{e}^{x^2}{\displaystyle \frac{\cos e^{x^2}}{\mathrm{sen}{e}^{x^2}}}$

$\mathbf{\text{c)}}$ $f(x)=\mathrm{arctg}(\mathrm{sen}(e^{\sqrt{x}}))$. Como $(e^{\sqrt{x}}{)}^{\mathrm{\prime }}=e^{\sqrt{x}}\cdot {\displaystyle \frac{1}{2\sqrt{x}}}$:

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{1}{1+{\mathrm{sen}}^2(e^{\sqrt{x}})}}\cdot \cos (e^{\sqrt{x}})\cdot {\displaystyle \frac{e^{\sqrt{x}}}{2\sqrt{x}}}={\displaystyle \frac{\cos (e^{\sqrt{x}})e^{\sqrt{x}}}{2\sqrt{x}[1+{\mathrm{sen}}^2(e^{\sqrt{x}})]}}$

$\mathbf{\text{d)}}$ $f(x)=[\mathrm{sen}({\cos }^3((4x^3+3{)}^2)){]}^2$. Sea $p=(4x^3+3{)}^2$, con $p^{\mathrm{\prime }}=2(4x^3+3)\cdot 12x^2=24x^2(4x^3+3)$. Derivando capa a capa:

$f^{\mathrm{\prime }}(x)=2\mathrm{sen}({\cos }^{3}p)\cdot \cos ({\cos }^{3}p)\cdot 3{\cos }^{2}p\cdot (-\mathrm{sen}p)\cdot 24x^2(4x^3+3)$

Agrupando las constantes $(2\cdot 3\cdot 24=144)$:

$f^{\mathrm{\prime }}(x)=-144x^2(4x^3+3)\mathrm{sen}({\cos }^{3}p)\cos ({\cos }^{3}p){\cos }^{2}p\mathrm{sen}p$, con $p=(4x^3+3{)}^2$.

$\mathbf{\text{e)}}$ $f(x)=2[\mathrm{tg}({\ln }^3((x^4+7{)}^5)){]}^2$. Conviene simplificar el interior: como $\ln ((x^4+7{)}^5)=5\ln (x^4+7)$, llamamos $L={\ln }^3((x^4+7{)}^5)=(5\ln (x^4+7){)}^3=125(\ln (x^4+7){)}^3$.

Su derivada:

$L^{\mathrm{\prime }}=125\cdot 3(\ln (x^4+7){)}^2\cdot {\displaystyle \frac{4x^3}{x^4+7}}={\displaystyle \frac{1500x^3(\ln (x^4+7){)}^2}{x^4+7}}$

Como $f=2(\mathrm{tg}L{)}^2$, entonces $f^{\mathrm{\prime }}=2\cdot 2\mathrm{tg}(L)\cdot {\displaystyle \frac{1}{{\cos }^{2}L}}\cdot L^{\mathrm{\prime }}$, es decir:

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{4\mathrm{tg}L}{{\cos }^{2}L}}\cdot {\displaystyle \frac{1500x^3(\ln (x^4+7){)}^2}{x^4+7}}$, con $L=125(\ln (x^4+7){)}^3$.

$\mathbf{\text{f)}}$ $f(x)={\ln }^3\sqrt[3]{{\mathrm{arctg}}^4{e}^w}$, con $w={\displaystyle \frac{x^2+\ln x}{x^2}}=1+{\displaystyle \frac{\ln x}{x^2}}$.

Primero la derivada del exponente:

$w^{\mathrm{\prime }}={\displaystyle \frac{\frac{1}{x}\cdot x^2-\ln x\cdot 2x}{x^4}}={\displaystyle \frac{x-2x\ln x}{x^4}}={\displaystyle \frac{1-2\ln x}{x^3}}$

Conviene simplificar usando las propiedades del logaritmo. La raíz cúbica y la potencia cuarta salen del logaritmo:

$\ln \sqrt[3]{{\mathrm{arctg}}^4{e}^w}={\displaystyle \frac{1}{3}}\ln ({\mathrm{arctg}}^4{e}^w)={\displaystyle \frac{4}{3}}\ln (\mathrm{arctg}{e}^w)$

Llamando $B=\mathrm{arctg}{e}^w$, queda $f=({\displaystyle \frac{4}{3}}\ln B{)}^3={\displaystyle \frac{64}{27}}(\ln B{)}^3$.

La derivada de $B$ es $B^{\mathrm{\prime }}={\displaystyle \frac{1}{1+e^{2w}}}\cdot e^w\cdot w^{\mathrm{\prime }}={\displaystyle \frac{e^w{w}^{\mathrm{\prime }}}{1+e^{2w}}}$. Entonces:

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{64}{27}}\cdot 3(\ln B{)}^2\cdot {\displaystyle \frac{B^{\mathrm{\prime }}}{B}}={\displaystyle \frac{64}{9}}\cdot {\displaystyle \frac{(\ln B{)}^2}{B}}\cdot {\displaystyle \frac{e^w{w}^{\mathrm{\prime }}}{1+e^{2w}}}$

Sustituyendo $B=\mathrm{arctg}{e}^w$ y $w^{\mathrm{\prime }}={\displaystyle \frac{1-2\ln x}{x^3}}$:

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{64}{9}}\cdot {\displaystyle \frac{(\ln (\mathrm{arctg}{e}^w){)}^2}{\mathrm{arctg}{e}^w}}\cdot {\displaystyle \frac{e^w}{1+e^{2w}}}\cdot {\displaystyle \frac{1-2\ln x}{x^3}}$, con $w=1+{\displaystyle \frac{\ln x}{x^2}}$.