Tabla de derivadas, ejemplos y ejercicios con solución.

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	Función	Derivada
(1) Derivada de una contante por una función.	$F(x)=c\cdot f(x)$	$F^{\prime }(x)=c\cdot f{(}^{\prime }x)$
(2) Derivada de una suma o resta de funciones.	$F(x)=f(x)\pm g(x)$	$F^{\prime }(x)=f^{\prime }(x)\pm g^{\prime }(x)$
(3) Derivada de un producto de funciones.	$F(x)=f(x)\cdot g(x)$	$F(x)=f^{\prime }(x)\cdot g(x)+f(x)\cdot g^{\prime }(x)$
(4) Derivada de un cociente de funciones.	$F(x)={\displaystyle \frac{f(x)}{g(x)}}$	$F^{\prime }(x)={\displaystyle \frac{f^{\prime }(x)\cdot g(x)-f(x)\cdot g^{\prime }(x)}{{[g(x)]}^2}}$
(5) Regla de la cadena	$F(x)=f(g(x))$	$F^{\prime }(x)=f^{\prime }(g(x))\cdot g^{\prime }(x)$

Linealidad de la derivada:

De las propiedades (1) y (2) podemos deducir que:

$$\text{Sean }{c}_1,c_2\in \mathbb{R}\Rightarrow {[c_1\cdot f(x)+c_2\cdot g(x)]}^{\prime }=c_1\cdot f^{\prime }(x)+c_2\cdot g^{\prime }(x)$$

Función	Derivada	Función compuesta	Derivada	Ejemplo
$y=c,c\in \mathbb{R}$	$y^{\prime }=0$
$y=x$	$y^{\prime }=1$
$y=x^2$	$y^{\prime }=2x$
$y=x^p,p\in \mathbb{R}$	$y^{\prime }=p\cdot x^{p-1}$	$F(x)=[f(x){]}^p$	$F^{\prime }(x)=p\cdot [f(x){]}^{p-1}\cdot f^{\prime }(x)$	Ej. 1
$y=\sqrt{x}$	$y^{\prime }={\displaystyle \frac{1}{2\sqrt{x}}}$	$F(x)=\sqrt{f(x)}$	$F^{\prime }(x)={\displaystyle \frac{1}{2\sqrt{f(x)}}}\cdot f^{\prime }(x)=$ $={\displaystyle \frac{f^{\prime }(x)}{2\sqrt{f(x)}}}$	Ej. 2
$y=e^x$	$y^{\prime }=e^x$	$F(x)=e^{f(x)}$	$F^{\prime }(x)=e^{f(x)}\cdot f^{\prime }(x)$	Ej. 3
$y=a^x$	$y^{\prime }=a^x\cdot \ln a$	$F(x)=a^{f(x)}=$ $=e^{\ln a^{f(x)}}=$ $=e^{f(x)\cdot \ln a}$	$F^{\prime }(x)=a^{f(x)}\cdot \ln a\cdot f^{\prime }(x)$	Ej. 4
$y=\ln x$	$y^{\prime }={\displaystyle \frac{1}{x}}$	$F(x)=\ln f(x)$	$F^{\prime }(x)={\displaystyle \frac{1}{f(x)}}\cdot f^{\prime }(x)=$ $={\displaystyle \frac{f^{\prime }(x)}{f(x)}}$	Ej. 5
$y={\log }_{a}x$	$y^{\prime }={\displaystyle \frac{1}{x}}\cdot {\displaystyle \frac{1}{\ln a}}$	$F(x)={\log }_{a}f(x)$ $={\displaystyle \frac{\ln f(x)}{\ln a}}$	$F^{\prime }(x)={\displaystyle \frac{1}{f(x)}}\cdot {\displaystyle \frac{1}{\ln a}}\cdot f^{\prime }(x)=$ $={\displaystyle \frac{f^{\prime }(x)}{f(x)\ln a}}$	Ej. 6
$y=\mathrm{sen}x$	$y^{\prime }=\cos x$	$F(x)=\mathrm{sen}f(x)$	$F^{\prime }(x)=f^{\prime }(x)\cos f(x)$	Ej. 7
$y=\cos x$	$y^{\prime }=-\mathrm{sen}x$	$F(x)=\cos f(x)$	$F^{\prime }(x)=-f^{\prime }(x)\mathrm{sen}f(x)$	Ej. 8
$y=\mathrm{tg}x=$ $={\displaystyle \frac{\mathrm{sen}x}{\cos x}}$	$y^{\prime }=1+{\mathrm{tg}}^{2}x=$ $={\sec }^{2}x$	$F(x)=\mathrm{tg}f(x)$	$F^{\prime }(x)=f^{\prime }(x)[1+{\mathrm{tg}}^2(f(x))]=$ $=f^{\prime }(x)\cdot {\sec }^{2}f(x)$	Ej. 9
$y=\sec x=$ $={\displaystyle \frac{1}{\cos x}}$	$y^{\prime }=\mathrm{tg}x\cdot \sec x$	$F(x)=\sec f(x)$	$F^{\prime }(x)=f^{\prime }(x)\cdot \mathrm{tg}f(x)\cdot \sec f(x)$	Ej. 10
$y=\mathrm{cosec}x$ $={\displaystyle \frac{1}{\mathrm{sen}x}}$	$y^{\prime }=-\mathrm{cotg}x\mathrm{cosec}x$	$F(x)=\mathrm{cosec}f(x)$	$F^{\prime }(x)=-f^{\prime }(x)\cdot \mathrm{cotg}f(x)\cdot \mathrm{sen}f(x)$	Ej. 11
$y=\mathrm{cotg}x$ $={\displaystyle \frac{\cos x}{\mathrm{sen}x}}$	$y^{\prime }=-(1+{\mathrm{cotg}}^{2}x)=$ $={\mathrm{cosec}}^{2}x$	$F(x)=\mathrm{cotg}f(x)$	$F^{\prime }(x)=-f^{\prime }(x)[1+{\mathrm{cotg}}^2(f(x))]=$ $=-f^{\prime }(x)\cdot {\mathrm{cosec}}^{2}f(x)$	Ej. 12
$y=\mathrm{arcsen}x$	$y^{\prime }={\displaystyle \frac{1}{\sqrt{1-x^2}}}$	$F(x)=\mathrm{arcsen}f(x)$	$F^{\prime }(x)={\displaystyle \frac{f^{\prime }(x)}{\sqrt{1-[f(x){]}^2}}}$	Ej. 13
$y=\arccos x$	$y^{\prime }={\displaystyle \frac{-1}{\sqrt{1-x^2}}}$	$F(x)=\arccos f(x)$	$F^{\prime }(x)={\displaystyle \frac{-f^{\prime }(x)}{\sqrt{1-[f(x){]}^2}}}$	Ej. 14
$y=\mathrm{arctg}x$	$y^{\prime }={\displaystyle \frac{1}{1+x^2}}$	$F(x)=\mathrm{arctg}f(x)$	$F^{\prime }(x)={\displaystyle \frac{f^{\prime }(x)}{1+[f(x){]}^2}}$	Ej. 15
$y=\mathrm{arcsec}x$	$y^{\prime }={\displaystyle \frac{1}{x\cdot \sqrt{x^2-1}}}$	$F(x)=\mathrm{arcsec}f(x)$	$F^{\prime }(x)={\displaystyle \frac{f^{\prime }(x)}{f(x)\cdot \sqrt{[f(x){]}^2-1}}}$	Ej. 16
$y=\mathrm{arccosec}x$	$y^{\prime }={\displaystyle \frac{-1}{x\cdot \sqrt{x^2-1}}}$	$F(x)=\mathrm{arccosec}f(x)$	$F^{\prime }(x)={\displaystyle \frac{-f^{\prime }(x)}{f(x)\cdot \sqrt{[f(x){]}^2-1}}}$	Ej. 17
$y=\mathrm{arccotg}x$	$y^{\prime }={\displaystyle \frac{-1}{1+x^2}}$	$F(x)=\mathrm{arccotg}f(x)$	$F^{\prime }(x)={\displaystyle \frac{-f^{\prime }(x)}{1+[f(x){]}^2}}$	Ej. 18
$F(x)=g(x{)}^{h(x)}$	$F^{\prime }(x)=h(x)\cdot g(x{)}^{h(x)-1}{g}^{\prime }(x)+g(x{)}^{h(x)}\cdot h^{\prime }(x)\cdot \ln g(x)$			Ej. 19
$y=\mathrm{sh}x$	$y^{\prime }=\mathrm{ch}x$	$F(x)=\mathrm{sh}f(x)$	$F(x)=f^{\prime }(x)\cdot \mathrm{sh}f(x)$
$y=\mathrm{ch}x$	$y^{\prime }=\mathrm{sh}x$	$F(x)=\mathrm{ch}f(x)$	$F(x)=f^{\prime }(x)\cdot \mathrm{sh}(f(x))$
$y=\mathrm{th}x$	$y^{\prime }={\displaystyle \frac{1}{{\mathrm{ch}}^{2}x}}=$ $=1-{\mathrm{th}}^{2}x$	$F(x)=\mathrm{th}f(x)$	$F(x)=f^{\prime }(x)\cdot {\displaystyle \frac{1}{{\mathrm{ch}}^{2}f(x)}}=$ $=f^{\prime }(x)\cdot (1-{\mathrm{th}}^{2}f(x))$
$y=\mathrm{argsh}x$	$y^{\prime }={\displaystyle \frac{1}{\sqrt{x^2+1}}}$	$F(x)=\mathrm{argsh}f(x)$	$F^{\prime }(x)={\displaystyle \frac{f^{\prime }(x)}{\sqrt{[f(x){]}^2+1}}}$
$y=\mathrm{argch}x$	$y^{\prime }={\displaystyle \frac{1}{\sqrt{x^2-1}}}$	$F(x)=\mathrm{argch}f(x)$	$F^{\prime }(x)={\displaystyle \frac{f^{\prime }(x)}{\sqrt{[f(x){]}^2-1}}}$
$y=\mathrm{argth}x$	$y^{\prime }={\displaystyle \frac{1}{1-x^2}}$	$F(x)=\mathrm{argth}f(x)$	$F^{\prime }(x)={\displaystyle \frac{f^{\prime }(x)}{1-[f(x){]}^2}}$

Derivada de la potencia de una función

$$f(x)={\ln }^5(3x)={(\ln (3x))}^5\Rightarrow f^{\prime }(x)=5\cdot {\ln }^4(3x)\cdot 3=15\cdot {\ln }^4(3x)$$

$$g(x)={\mathrm{tg}}^3\left(e^{3x}\right)\Rightarrow g^{\prime }(x)=3\cdot {\mathrm{tg}}^2\left(e^{3x}\right)\cdot (1+{\mathrm{tg}}^2\left(e^{3x}\right))\cdot 3=9{\mathrm{tg}}^2(3x)(1+{\mathrm{tg}}^2(3x))$$

Derivada de la función raíz cuadrada

$$f(x)=\sqrt{{\displaystyle \frac{1-x}{1+x}}}\Rightarrow f^{\prime }(x)={\displaystyle \frac{{\left(\sqrt{{\displaystyle \frac{1-x}{1+x}}}\right)}^{\prime }}{2\cdot \sqrt{{\displaystyle \frac{1-x}{1+x}}}}}={\displaystyle \frac{{\displaystyle \frac{-(1+x)-(1-x)}{(1+x{)}^2}}}{2\cdot \sqrt{{\displaystyle \frac{1-x}{1+x}}}}}={\displaystyle \frac{{\displaystyle \frac{-1-x-1+x}{(1+x{)}^2}}}{2\cdot \sqrt{{\displaystyle \frac{1-x}{1+x}}}}}={\displaystyle \frac{{\displaystyle \frac{-2}{(1+x{)}^2}}}{2\cdot \sqrt{{\displaystyle \frac{1-x}{1+x}}}}}=$$

$$={\displaystyle \frac{{\displaystyle \frac{-1}{(1+x{)}^2}}}{\sqrt{{\displaystyle \frac{1-x}{1+x}}}}}={\displaystyle \frac{-\sqrt{1+x}}{(1+x{)}^2\sqrt{1-x}}}={\displaystyle \frac{-\sqrt{1-x^2}}{(1-x)(1+x{)}^2}}$$

$$g(x)=\sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}\Rightarrow g^{\prime }(x)={\displaystyle \frac{{\left({\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}\right)}^{\prime }}{2\cdot \sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}={\displaystyle \frac{{\displaystyle \frac{-\cos x(1+\mathrm{sen}x)-\cos x(1-\mathrm{sen}x)}{(1+\mathrm{sen}x{)}^2}}}{2\cdot \sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}=$$
$$={\displaystyle \frac{{\displaystyle \frac{-\cos x-\cos x\mathrm{sen}x-\cos x+\cos x\mathrm{sen}x}{(1+\mathrm{sen}x{)}^2}}}{2\cdot \sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}={\displaystyle \frac{{\displaystyle \frac{-2\cos x}{(1+\mathrm{sen}x{)}^2}}}{2\cdot \sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}={\displaystyle \frac{{\displaystyle \frac{-\cos x}{(1+\mathrm{sen}x{)}^2}}}{\sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}={\displaystyle \frac{-\cos x\sqrt{1+\mathrm{sen}x}}{(1+\mathrm{sen}x{)}^2\cdot \sqrt{1-\mathrm{sen}x}}}=$$
$$={\displaystyle \frac{-\cos x\sqrt{1-{\mathrm{sen}}^{2}x}}{(1+\mathrm{sen}x{)}^2\cdot (1-\mathrm{sen}x)}}={\displaystyle \frac{-{\cos }^{2}x}{(1+\mathrm{sen}x)\cdot (1-{\mathrm{sen}}^{2}x)}}={\displaystyle \frac{-{\cos }^{2}x}{(1+\mathrm{sen}x)\cdot {\cos }^{2}x}}={\displaystyle \frac{-1}{1+\mathrm{sen}x}}$$

Derivada de la función exponencial, base $e$

$$f(x)=e^{\mathrm{sen}x}=\Rightarrow f^{\prime }(x)=\cos x\cdot e^{\mathrm{sen}x}$$

$$g(x)=e^{x\cdot \cos x}\Rightarrow g^{\prime }(x)=(x\cdot \cos x{)}^{\prime }\cdot e^{x\cdot \cos x}=(\cos x-x\mathrm{sen}x)\cdot e^{x\cdot \cos x}$$

Derivada de la función exponencial, base $a\ne e$

$$f(x)=2^x=e^{\ln 2^x}=e^{x\cdot \ln 2}\Rightarrow f^{\prime }(x)=\ln 2\cdot e^{x\cdot \ln 2}=\ln 2\cdot 2^x$$

Derivada de la función logaritmo neperiano

Para este tipo de derivadas hay que tener en cuenta las propiedades de los logartimos:

$$f(x)=\ln (x+\sqrt{1+x^2})\Rightarrow f^{\prime }(x)={\displaystyle \frac{(x+\sqrt{1+x^2}{)}^{\prime }}{x+\sqrt{1+x^2}}}={\displaystyle \frac{1+{\displaystyle \frac{2x}{2\cdot (\sqrt{1+x^2})}}}{x+\sqrt{1+x^2}}}={\displaystyle \frac{1+{\displaystyle \frac{x}{\sqrt{1+x^2}}}}{x+\sqrt{1+x^2}}}={\displaystyle \frac{{\displaystyle \frac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}}}}{x+\sqrt{1+x^2}}}=$$
$$={\displaystyle \frac{{\displaystyle \frac{\cancel{x+\sqrt{1+x^2}}}{\sqrt{1+x^2}}}}{\cancel{x+\sqrt{1+x^2}}}}={\displaystyle \frac{1}{\sqrt{1+x^2}}}={\displaystyle \frac{\sqrt{1+x^2}}{1+x^2}}$$

$$g(x)=\ln \left({\displaystyle \frac{\sqrt{a^2+x^2}}{x^2}}\right)=\ln \left(\sqrt{a^2+x^2}\right)-\ln x^2={\displaystyle \frac{1}{2}}\ln (a^2+x^2)-2\ln x\Rightarrow f^{\prime }(x)={\displaystyle \frac{1}{2}}{\displaystyle \frac{2x}{a^2+x^2}}-{\displaystyle \frac{2}{x}}=$$
$$={\displaystyle \frac{x}{a^2+x^2}}-{\displaystyle \frac{2}{x}}={\displaystyle \frac{x^2-2a^2-2x^2}{x(a^2+x^2)}}={\displaystyle \frac{-(x^2+2a^2)}{x(a^2+x^2)}}$$

$$h(x)=\ln \sqrt{{\displaystyle \frac{1-x}{1+x}}}=\ln {\left({\displaystyle \frac{1-x}{1+x}}\right)}^{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{2}}\cdot [\ln (1-x)-\ln (1+x)]\Rightarrow$$ $$\Rightarrow h^{\prime }(x)={\displaystyle \frac{1}{2}}\cdot [{\displaystyle \frac{-1}{1-x}}-{\displaystyle \frac{1}{1+x}}]={\displaystyle \frac{1}{2}}\cdot [{\displaystyle \frac{-1-x}{(1-x)(1+x)}}-{\displaystyle \frac{1-x}{(1-x)(1+x)}}]={\displaystyle \frac{1}{2}}\cdot \left[{\displaystyle \frac{-1-x-1+x}{(1-x)(1+x)}}\right]=$$ $$={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{-2}{(1-x)(1+x)}}={\displaystyle \frac{1}{\cancel{2}}}\cdot {\displaystyle \frac{-\cancel{2}}{(1-x)(1+x)}}={\displaystyle \frac{-1}{1-x^2}}={\displaystyle \frac{1}{x^2-1}}$$

$$i(x)=\ln {(x\cdot \mathrm{tg}x)}^2=2\cdot \ln (x\cdot \mathrm{tg}x)=2\cdot (\ln x+\ln \mathrm{tg}x)$$ $$i^{\prime }(x)=2\cdot ({\displaystyle \frac{1}{x}}+{\displaystyle \frac{1+{\mathrm{tg}}^{2}x}{\mathrm{tg}x}})={\displaystyle \frac{2}{x}}+2\mathrm{cotg}x+2\mathrm{tg}x$$

$$j(x)=\ln ({\displaystyle \frac{1}{x^2}}\cdot \sqrt[3]{x^2-1})=\ln \left({\displaystyle \frac{1}{x^2}}\right)+\ln \left(\sqrt[3]{x^2-1}\right)=\ln x^{-2}+\ln {(x^2-1)}^{1/3}=-2\ln x+{\displaystyle \frac{1}{3}}\cdot \ln (x^2-1)$$ $$j^{\prime }(x)={\displaystyle \frac{-2}{x}}+{\displaystyle \frac{1}{3}}\cdot {\displaystyle \frac{2x}{x^2-1}}={\displaystyle \frac{-6x^2+6+2x^2}{3x(x^2-1)}}={\displaystyle \frac{6-4x^2}{3x(x^2-1)}}$$

$$k(x)=\ln \left(\sqrt[3]{{\displaystyle \frac{1}{(1+x{)}^2}}}\right)=\ln {\left({\displaystyle \frac{1}{(1+x{)}^2}}\right)}^{1/3}=\ln (1+x{)}^{-2/3}={\displaystyle \frac{-2}{3}}\cdot \ln (1+x)$$ $$k^{\prime }(x)={\displaystyle \frac{-2}{3}}\cdot {\displaystyle \frac{1}{1+x}}={\displaystyle \frac{-2}{3+3x}}$$

Derivada de la función logaritmo en base $a\ne e$

Hacemos un cambio de base y tenemos que derivar un logaritmo neperiano cuya derivada ya hemos visto: $$F(x)={\log }_{a}f(x)={\displaystyle \frac{\ln f(x)}{\ln a}}\Rightarrow F^{\prime }(x)={\displaystyle \frac{1}{\ln a}}\cdot {\displaystyle \frac{f^{\prime }(x)}{f(x)}}$$

$$g(x)={\log }_3(x+2)={\displaystyle \frac{\ln (x+2)}{\ln 3}}\Rightarrow g^{\prime }(x)={\displaystyle \frac{1}{\ln 3\cdot (x+2)}}$$

Derivada de la función seno

$$f(x)=\mathrm{sen}\left(e^{3x}\right)\Rightarrow f^{\prime }(x)=(e^{3x}{)}^{\prime }\cdot \cos \left(e^{3x}\right)=3\cdot e^{3x}\cdot \cos \left(e^{3x}\right)$$

$$g(x)=\mathrm{sen}(\ln x)\Rightarrow g^{\prime }(x)=(\ln x{)}^{\prime }\cdot \cos (\ln x)={\displaystyle \frac{1}{x}}\cdot \cos (\ln x)={\displaystyle \frac{\cos (\ln x)}{x}}$$

$$h(x)=\mathrm{sen}(x\cdot \mathrm{sen}x)\Rightarrow h^{\prime }(x)=(x\cdot \mathrm{sen}x{)}^{\prime }\cos (x\cdot \mathrm{sen}x)=(\mathrm{sen}x+x\cdot \cos x)\cos (x\cdot \mathrm{sen}x)$$

Derivada de la función coseno

$$f(x)=\cos \left(e^{5x}\right)\Rightarrow f^{\prime }(x)=-5\cdot e^{5x}\cdot \mathrm{sen}\left(e^{5x}\right)$$

$$g(x)=\cos \left({\displaystyle \frac{e^x}{5x}}\right)\Rightarrow g^{\prime }(x)=-{\left({\displaystyle \frac{e^x}{5x}}\right)}^{\prime }\cdot \mathrm{sen}\left({\displaystyle \frac{e^x}{5x}}\right)=-{\displaystyle \frac{e^x\cdot 5x-5\cdot e^x}{25x^2}}\cdot \mathrm{sen}\left({\displaystyle \frac{e^x}{5x}}\right)=$$
$$={\displaystyle \frac{5e^x(1-x)}{25x^2}}\cdot \mathrm{sen}\left({\displaystyle \frac{e^x}{5x}}\right)={\displaystyle \frac{e^x(1-x)}{5x^2}}\cdot \mathrm{sen}\left({\displaystyle \frac{e^x}{5x}}\right)$$

Derivada de la función exponencial, base distinta del número $e$

Tenemos que derivar $2^x$, si aplicamos las propiedades de los logaritmos vemos que $2^x=e^{\ln 2^x}=e^{x\ln 2}$ que ya sabemos derivar. $$f(x)=\mathrm{tg}{2}^x=\mathrm{tg}{e}^{\ln 2^x}=\mathrm{tg}{e}^{x\ln 2}\Rightarrow f^{\prime }(x)=(2^x{)}^{\prime }\cdot (1+{\mathrm{tg}}^2{2}^x)=\ln 2\cdot 2^x\cdot (1+{\mathrm{tg}}^2{2}^x)$$

$$g(x)=\mathrm{tg}\left({\displaystyle \frac{1-e^x}{1+e^x}}\right)\Rightarrow g^{\prime }(x)={\left({\displaystyle \frac{1-e^x}{1+e^x}}\right)}^{\prime }\cdot (1+{\mathrm{tg}}^2\left({\displaystyle \frac{1-e^x}{1+e^x}}\right))=$$
$$=\left({\displaystyle \frac{-e^x(1+e^x)-e^x(1-e^x)}{1+e^x}}\right)\cdot (1+{\mathrm{tg}}^2\left({\displaystyle \frac{1-e^x}{1+e^x}}\right))=\left({\displaystyle \frac{-e^x-e^{2x}-e^x+e^{2x}}{{(1+e^x)}^2}}\right)\cdot (1+{\mathrm{tg}}^2\left({\displaystyle \frac{1-e^x}{1+e^x}}\right))=$$
$$={\displaystyle \frac{-2e^x}{{(1+e^x)}^2}}\cdot (1+{\mathrm{tg}}^2\left({\displaystyle \frac{1-e^x}{1+e^x}}\right))$$

Derivada de la función arco secante

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Derivada de la función arco cosecante

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Derivada de la función arco cotangente

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Derivada de la función arco seno

$$f(x)=\mathrm{arcsen}(x^2-4x)\Rightarrow f^{\prime }(x)={\displaystyle \frac{2x-4}{\sqrt{1-(x^2-4x{)}^2}}}$$ $$g(x)=\mathrm{arcsen}({\log }_5(3x))=\mathrm{arcsen}\left({\displaystyle \frac{\ln 3x}{\ln 5}}\right)\Rightarrow g^{\prime }(x)={\displaystyle \frac{{\displaystyle \frac{1}{\ln 5}}\cdot {\displaystyle \frac{3}{3x}}}{\sqrt{1-({\log }_5(3x){)}^2}}}={\displaystyle \frac{1}{x\cdot \ln 5\cdot \sqrt{1-({\log }_5(3x){)}^2}}}$$

Derivada de la función arco coseno

$$h(x)=\arccos (\ln x)\Rightarrow h^{\prime }(x)=-{\displaystyle \frac{{\displaystyle \frac{1}{x}}}{\sqrt{1-(\ln x{)}^2}}}={\displaystyle \frac{-1}{x\cdot \sqrt{1-{\ln }^{2}x}}}$$ $$$$

$$f(x)=\arccos \left({\displaystyle \frac{x}{\sqrt{1+x^2}}}\right)\Rightarrow f^{\prime }(x)={\displaystyle \frac{{\left({\displaystyle \frac{x}{\sqrt{1+x^2}}}\right)}^{\prime }}{\sqrt{1-{\left({\displaystyle \frac{x}{\sqrt{1+x^2}}}\right)}^2}}}={\displaystyle \frac{{\displaystyle \frac{\sqrt{1+x^2}-x{\displaystyle \frac{2x}{2\cdot \sqrt{1+x^2}}}}{1+x^2}}}{\sqrt{1-{\displaystyle \frac{x^2}{1+x^2}}}}}=$$ $$={\displaystyle \frac{{\displaystyle \frac{\sqrt{1+x^2}-x{\displaystyle \frac{x}{\sqrt{1+x^2}}}}{1+x^2}}}{\sqrt{{\displaystyle \frac{1+x^2-x^2}{1+x^2}}}}}={\displaystyle \frac{{\displaystyle \frac{{\displaystyle \frac{1+x^2-x^2}{\sqrt{1+x^2}}}}{1+x^2}}}{{\displaystyle \frac{1}{\sqrt{1+x^2}}}}}={\displaystyle \frac{{\displaystyle \frac{{\displaystyle \frac{1}{\sqrt{1+x^2}}}}{1+x^2}}}{{\displaystyle \frac{1}{\sqrt{1+x^2}}}}}={\displaystyle \frac{{\displaystyle \frac{1}{\sqrt{1+x^2}(1+x^2)}}}{{\displaystyle \frac{1}{\sqrt{1+x^2}}}}}={\displaystyle \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}(1+x^2)}}=$$
$$={\displaystyle \frac{1}{1+x^2}}$$

Derivada de la función arco tangente

$$F(x)=\mathrm{arccotg}\left(x^2\right)⟶F^{\prime }(x)={\displaystyle \frac{-2x}{1+{\left(x^2\right)}^2}}={\displaystyle \frac{-2x}{1+x^4}}$$

$$g(x)=\mathrm{arctg}\sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}\Rightarrow g^{\prime }(x)={\displaystyle \frac{{\left(\sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}\right)}^{\prime }}{1+{\left(\sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}\right)}^2}}={\displaystyle \frac{{\displaystyle \frac{{\left({\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}\right)}^{\prime }}{2\cdot \sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}}{1+{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}={\displaystyle \frac{{\displaystyle \frac{{\left({\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}\right)}^{\prime }}{2\cdot \sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}}{{\displaystyle \frac{1+\mathrm{sen}x+1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}=$$
$$={\displaystyle \frac{{\displaystyle \frac{{\left({\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}\right)}^{\prime }}{2\cdot \sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}}{{\displaystyle \frac{2}{1+\mathrm{sen}x}}}}={\displaystyle \frac{{\displaystyle \frac{{\displaystyle \frac{-\cos x(1+\mathrm{sen}x)-\cos x(1-\mathrm{sen}x)}{(1+\mathrm{sen}x{)}^2}}}{2\cdot \sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}}{{\displaystyle \frac{2}{1+\mathrm{sen}x}}}}={\displaystyle \frac{{\displaystyle \frac{{\displaystyle \frac{-\cos x+\cos x\mathrm{sen}x-\cos x+\cos x\mathrm{sen}x}{(1+\mathrm{sen}x{)}^2}}}{2\cdot \sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}}{{\displaystyle \frac{2}{1+\mathrm{sen}x}}}}=$$
$$={\displaystyle \frac{{\displaystyle \frac{{\displaystyle \frac{-2\cos x}{(1+\mathrm{sen}x{)}^2}}}{2\cdot \sqrt{{\displaystyle \frac{1-\mathrm{sen}x}{1+\mathrm{sen}x}}}}}}{{\displaystyle \frac{2}{1+\mathrm{sen}x}}}}={\displaystyle \frac{{\displaystyle \frac{-2\cos x(\sqrt{1+\mathrm{sen}x})}{2\cdot \sqrt{1-\mathrm{sen}x}(1+\mathrm{sen}x{)}^2}}}{{\displaystyle \frac{2}{1+\mathrm{sen}x}}}}={\displaystyle \frac{{\displaystyle \frac{-\cos x(\sqrt{1+\mathrm{sen}x})}{\sqrt{1-\mathrm{sen}x}(1+\mathrm{sen}x{)}^2}}}{{\displaystyle \frac{2}{1+\mathrm{sen}x}}}}=$$
$$={\displaystyle \frac{-\cos x(\sqrt{1+\mathrm{sen}x})(1+\mathrm{sen}x)}{2\cdot \sqrt{1-\mathrm{sen}x}(1+\mathrm{sen}x{)}^2}}={\displaystyle \frac{-\cos x(\sqrt{1+\mathrm{sen}x})\cancel{(1+\mathrm{sen}x)}}{2\cdot \sqrt{1-\mathrm{sen}x}(1+\mathrm{sen}x{)}^{\cancel{2}}}}={\displaystyle \frac{-\cos x(\sqrt{1+\mathrm{sen}x})}{2\cdot \sqrt{1-\mathrm{sen}x}(1+\mathrm{sen}x)}}=$$
$$={\displaystyle \frac{-\cos x(\sqrt{1-{\mathrm{sen}}^{2}x})}{2\cdot (1-\mathrm{sen}x)(1+\mathrm{sen}x)}}={\displaystyle \frac{-\cos x(\sqrt{1-{\mathrm{sen}}^{2}x})}{2\cdot (1-\mathrm{sen}x)(1+\mathrm{sen}x)}}={\displaystyle \frac{-\cos x(\sqrt{{\cos }^{2}x})}{2\cdot (1-{\mathrm{sen}}^{2}x)}}={\displaystyle \frac{-{\cos }^{2}x}{2\cdot {\cos }^{2}x}}={\displaystyle \frac{-1}{2}}$$

Derivada de la función arco secante

$$f(x)=\mathrm{arcsec}(x^7)\Rightarrow f^{\prime }(x)={\displaystyle \frac{7x^6}{x^7\cdot \sqrt{{\left(x^7\right)}^2-1}}}={\displaystyle \frac{7}{x\cdot \sqrt{x^{14}-1}}}$$

$$g(x)=(\mathrm{arcsec}x{)}^9\Rightarrow f^{\prime }(x)=9\cdot (\mathrm{arcsec}x{)}^8\cdot (\mathrm{arcsec}x{)}^{\prime }=9\cdot (\mathrm{arcsec}x{)}^8\cdot {\displaystyle \frac{1}{x\cdot \sqrt{x^2-1}}}$$

Derivada de la función arco cosecante

$$f(x)=\mathrm{arccosec}\left({\displaystyle \frac{1+x}{1-x}}\right)\Rightarrow f^{\prime }(x)={\displaystyle \frac{-{\displaystyle \frac{1-x+1+x}{(1-x{)}^2}}}{\left({\displaystyle \frac{1+x}{1-x}}\right)\cdot \sqrt{{\left({\displaystyle \frac{1+x}{1-x}}\right)}^2-1}}}=$$
$$={\displaystyle \frac{-(1-x)+(1+x)}{(1-x)\cdot (1+x)\cdot \sqrt{{\left({\displaystyle \frac{1+x}{1-x}}\right)}^2-1}}}={\displaystyle \frac{-2x}{(1-x)\cdot (1+x)\cdot \sqrt{{\left({\displaystyle \frac{1+x}{1-x}}\right)}^2-1}}}={\displaystyle \frac{2x}{(1-x)\cdot (1+x)\cdot \sqrt{{\displaystyle \frac{(1+x{)}^2-(1-x{)}^2}{(1-x{)}^2}}}}}=$$
$$={\displaystyle \frac{2x}{(1+x)\cdot \sqrt{(1+x{)}^2-(1-x{)}^2}}}={\displaystyle \frac{2x}{(1+x)\cdot \sqrt{(1+2x+x^2)-(1-2x+x^2)}}}=$$
$$={\displaystyle \frac{2x}{(1+x)\cdot \sqrt{4x}}}={\displaystyle \frac{x}{(1+x)\cdot \sqrt{x}}}$$

$$g(x)=\mathrm{arccosec}({\log }_{5}9x)=\mathrm{arccosec}\left({\displaystyle \frac{\ln 9x}{\ln 5}}\right)\Rightarrow g^{\prime }(x)={\displaystyle \frac{-{\displaystyle \frac{1}{\ln 5}}\cdot {\displaystyle \frac{9}{9x}}}{({\log }_{5}9x)\cdot \sqrt{{({\log }_{5}9x)}^2-1}}}=$$
$$={\displaystyle \frac{-1}{\ln 5\cdot x\cdot ({\log }_{5}9x)\cdot \sqrt{{({\log }_{5}9x)}^2-1}}}$$

Derivada de la función arco cotangente

$$f(x)=\mathrm{arccotg}(6x^8+8x^7+3x)\Rightarrow f^{\prime }(x)={\displaystyle \frac{-(48x^7+56x^6+3)}{1+{(8x^7+6x^8+3x)}^2}}$$

Derivada de la función potencial-exponencial

Derivación logarítmica: Cuando tenemos una función potencial-exponencial, es decir, que tanto la base como el exponente son dos funciones no constantes, $F(x)=g(x{)}^{h(x)}$ y la queremos derivar. Tenemos dos opciones:

1. Aprendernos esta fórmula de memoria $F^{\prime }(x)=h(x)\cdot g(x{)}^{h(x)-1}{g}^{\prime }(x)+g(x{)}^{h(x)}\cdot h^{\prime }(x)\cdot \ln g(x)$
2. o saber deducirla (muy recomendable). Veamos como:
   
   Lo $1^{\underset{―}{o}}$ tomamos logaritmos neperianos: $$\ln F(x)=\ln g(x{)}^{h(x)}$$ $$\text{ usamos las propiedades de los logaritmos y nos queda }\ln F(x)=h(x)\cdot \ln g(x)$$ Tenemos dos expresiones iguales, luego sus funciones derivadas serán iguales: $${(\ln F(x))}^{\prime }={(h(x)\cdot \ln g(x))}^{\prime }\Rightarrow {\displaystyle \frac{F^{\prime }(x)}{F(x)}}=h^{\prime }(x)\ln g(x)+h(x)\cdot {\displaystyle \frac{g^{\prime }(x)}{g(x)}}$$ $$\text{ pasamos }F(x)\text{ multiplicando y tenemos que }{F}^{\prime }(x)=F(x)\cdot (h^{\prime }(x)\ln g(x)+h(x)\cdot {\displaystyle \frac{g^{\prime }(x)}{g(x)}})$$ y sustituyendo $F(x)$ por su valor tenemos: $$F^{\prime }(x)=g(x{)}^{h(x)}\cdot (h^{\prime }(x)\ln g(x)+h(x)\cdot {\displaystyle \frac{g^{\prime }(x)}{g(x)}})=g(x{)}^{h(x)}\cdot h^{\prime }(x)\cdot \ln g(x)+g(x{)}^{h(x)}\cdot h(x)\cdot {\displaystyle \frac{g^{\prime }(x)}{g(x)}}$$ y así tenemos: $$F^{\prime }(x)=g(x{)}^{h(x)}\cdot h^{\prime }(x)\cdot \ln g(x)+g(x{)}^{h(x)-1}\cdot h(x)\cdot g^{\prime }(x)$$
Vamos con un par de ejemplos: $$F(x)=(\mathrm{tg}x{)}^{\mathrm{cotg}x}$$ Tomamos logaritmos, aplicamos la propiedad de las potencias y derivamos: $$\ln F(x)=\mathrm{cotg}x\cdot \ln \mathrm{tg}x\Rightarrow {\displaystyle \frac{F^{\prime }(x)}{F(x)}}=-(1+{\mathrm{cotg}}^{2}x)\ln \mathrm{tg}x+\mathrm{cotg}x\cdot {\displaystyle \frac{1+{\mathrm{tg}}^{2}x}{\mathrm{tg}x}}$$ Pasamos $F(x)$ multiplicando y la sustituimos por su valor: $$F^{\prime }(x)=(\mathrm{tg}x{)}^{\mathrm{cotg}x}\cdot (-(1+{\mathrm{cotg}}^{2}x)\ln \mathrm{tg}x+\mathrm{cotg}x\cdot {\displaystyle \frac{1+{\mathrm{tg}}^{2}x}{\mathrm{tg}x}})=$$

$$=-(\mathrm{tg}x{)}^{\mathrm{cotg}x}\cdot (1+{\mathrm{cotg}}^{2}x)\ln \mathrm{tg}x+(\mathrm{tg}x{)}^{\mathrm{cotg}x}\cdot \mathrm{cotg}x\cdot {\displaystyle \frac{1+{\mathrm{tg}}^{2}x}{\mathrm{tg}x}}=(\ast )$$ Vemos que $$\mathrm{cotg}x\cdot {\displaystyle \frac{1+{\mathrm{tg}}^{2}x}{\mathrm{tg}x}}={\mathrm{cotg}}^{2}x\cdot (1+{\mathrm{tg}}^{2}x)={\mathrm{cotg}}^{2}x+1$$ Sustituyendo en (*) tenemos y sacando factor común: $$=-(\mathrm{tg}x{)}^{\mathrm{cotg}x}\cdot (1+{\mathrm{cotg}}^{2}x)\ln \mathrm{tg}x+(\mathrm{tg}x{)}^{\mathrm{cotg}x}\cdot (1+{\mathrm{cotg}}^{2}x)=-(\mathrm{tg}x{)}^{\mathrm{cotg}x}\cdot (1+{\mathrm{cotg}}^{2}x)\cdot (\ln \mathrm{tg}x-1)$$

Vamos con otro ejemplo: $$G(x)=\sqrt[\mathrm{tg}x]{x}=x^{{\displaystyle \frac{1}{\mathrm{tg}x}}}=x^{\mathrm{cotg}x}$$ Tomamos logaritmos, aplicamos la propiedad de las potencias y derivamos: $$\ln G(x)=\mathrm{cotg}x\cdot \ln x\Rightarrow {\displaystyle \frac{G^{\prime }(x)}{G(x)}}=-(1+{\mathrm{cotg}}^{2}x)\ln x+\mathrm{cotg}x\cdot {\displaystyle \frac{1}{x}}$$ Pasamos $G(x)$ multiplicando y la sustituimos por su valor: $$G^{\prime }(x)=\sqrt[\mathrm{tg}x]{x}\cdot (-(1+{\mathrm{cotg}}^{2}x)\ln x+\mathrm{cotg}x\cdot {\displaystyle \frac{1}{x}})=-\sqrt[\mathrm{tg}x]{x}\cdot (1+{\mathrm{cotg}}^{2}x)\cdot \ln x+\sqrt[\mathrm{tg}x]{x}\cdot \mathrm{cotg}x\cdot {\displaystyle \frac{1}{x}}$$ Agrupando nos queda: $$G^{\prime }(x)=\sqrt[\mathrm{tg}x]{x}\cdot (1+{\mathrm{cotg}}^{2}x)\cdot \ln x+\sqrt[\mathrm{cotg}x-1]{x}\cdot \mathrm{cotg}x$$ Otra variante $$G^{\prime }(x)=\sqrt[\mathrm{tg}x]{x}\cdot ({\displaystyle \frac{1}{x\cdot \mathrm{tg}x}}-{\displaystyle \frac{\ln x}{{\mathrm{sen}}^{2}x}})$$

Iremos actualizando esta sección con los ejercicios que nos pidáis, para ello podéis mandar un correo a profesor.maties@gmail.com