Derivación de funciones hiperbólicas

Proporcionamos ejercicios sobre derivación de funciones hiperbólicas.

Enunciado
1. Hallar las derivadas de las funciones:
   $(a)\; y=2x\operatorname{senh} x.\;\; (b)\;y=\dfrac{3x^2}{\cosh x}.\,\, (c)\;y=x-\tanh x.$
2. Demostrar que: $$\dfrac{d}{dx}\operatorname{senh} x =\cosh x,\quad\dfrac{d}{dx}\cosh x =\operatorname{senh} x,\quad\dfrac{d}{dx}\tanh x =\operatorname{sech}^2 x.$$
3. Calcular: $(a)\;\dfrac{d}{dx}\operatorname{csch} x.\;(b)\;\dfrac{d}{dx}\operatorname{sech} x.\;(c)\;\dfrac{d}{dx}\coth x. $

Solución
1. $(a)$ $y’=2\operatorname{senh} x+2x\cosh x=2(\operatorname{senh} x+x\cosh x).$
   $(b)$ $y’=\dfrac{6x\cosh x-3x^2\operatorname{senh} x}{\cosh^2x}=\dfrac{3(2x\cosh x-3x^2\operatorname{senh} x)}{\cosh^2x}.$
   $(c)$ $y’=1-\operatorname{sech}^2x=1-\dfrac{1}{\cosh^2x}=\dfrac{\cosh^2x-1}{\cosh^2x}=\dfrac{\operatorname{senh}^2 x}{\cosh^2x}=\tanh^2x.$
2. $$\dfrac{d}{dx}\operatorname{senh} x=\dfrac{d}{dx}\left(\dfrac{1}{2}\left(e^x-e^{-x}\right)\right)=\dfrac{1}{2}\dfrac{d}{dx}\left(e^x-\dfrac{1}{e^x}\right)=\dfrac{1}{2}\left(e^x-\dfrac{-e^x}{e^{2x}}\right)$$$$=\dfrac{1}{2}\left(e^x+e^{-x}\right)=\cosh x.$$$$\dfrac{d}{dx}\cosh x=\dfrac{d}{dx}\left(\dfrac{1}{2}\left(e^x+e^{-x}\right)\right)=\dfrac{1}{2}\dfrac{d}{dx}\left(e^x+\dfrac{1}{e^x}\right)=\dfrac{1}{2}\left(e^x+\dfrac{-e^x}{e^{2x}}\right)$$$$=\dfrac{1}{2}\left(e^x-e^{-x}\right)=\operatorname{senh} x.$$Nota. Se pueden acortar algo las demostraciones anteriores usando el teorema de derivación de funciones compuestas.$$\dfrac{d}{dx}\tanh x =\dfrac{d}{dx}\left(\dfrac{\operatorname{senh} x}{\cosh x}\right)=\dfrac{\cosh x\cosh x-\operatorname{senh} x \operatorname{senh} x}{\cosh^2 x}$$$$=\dfrac{\cosh^2 x-\operatorname{senh}^2 x}{\cosh^2 x}=\dfrac{1}{\cosh^2 x}=\operatorname{sech}^2 x.$$
3. $(a)$ $\dfrac{d}{dx}\operatorname{csch} x=\dfrac{d}{dx}\left(\dfrac{1}{\operatorname{senh} x}\right)=\dfrac{-\cosh x}{\operatorname{senh}^2 x}=-\coth x \operatorname{csch} x.$
   $(b)$ $\dfrac{d}{dx}\operatorname{sech} x=\dfrac{d}{dx}\left(\dfrac{1}{\cosh x}\right)=\dfrac{-\operatorname{senh} x}{\cosh^2 x}=-\tanh x \operatorname{sech} x.$
   $$(c)\qquad \dfrac{d}{dx}\coth x=\dfrac{d}{dx}\left(\dfrac{\cosh x}{\operatorname{senh} x}\right)=\dfrac{\operatorname{senh} x \operatorname{senh} x-\cosh x\cosh x}{\operatorname{senh}^2 x}$$$$=\dfrac{\operatorname{senh}^2 x-\cosh^2 x}{\operatorname{senh}^2 x}=\dfrac{-1}{\operatorname{senh}^2 x}=-\operatorname{csch}^2 x.$$