Derivadas: compendio de reglas y fórmulas

RESUMEN TEÓRICO
    Enunciado
  1. Hallar la derivada de $f(x)=\arccos \dfrac{1-x^2}{1+x^2}.$
  2. Hallar la derivada de $f(x)=\log \sqrt{\dfrac{\tan x+1}{\tan x-1}}.$
  3. Hallar la derivada de $f(x)=\operatorname{senh} \left(x\sqrt{x\sqrt{x}}\right).$
  4. Calcular las derivadas de: $(a)\;f(x)=\operatorname{arsh}\dfrac{x^2}{a^2}.\; (b)\;g(x)=\operatorname{arth}\dfrac{2x}{x^2+1}.$
  5. Demostrar las fórmulas de derivación de las funciones hiperbólicas inversas, es decir: $$\dfrac{d}{dx} \operatorname{arsh}x =\dfrac{1}{\sqrt{x^{2}+1}},\quad \dfrac{d}{dx} \operatorname{arch}\,x =\dfrac{1}{\sqrt{x^{2}-1}},\quad \dfrac{d}{dx} \operatorname{arth}\,x =\dfrac{1}{1-x^{2}}.$$
  6. Demostrar que $\dfrac{d}{dx}\left(\log \sqrt{\dfrac{1+\operatorname{sen} x}{1-\operatorname{sen} x}}\right)=\sec x.$
  7. Calcular la derivada de $y=\dfrac{x}{2}\sqrt{x^2+a}+\dfrac{a}{2}\log \left(x+\sqrt{x^2+a}\right).$
  8. Calcular $f’(4)$ siendo $f(x)=\sqrt{\sqrt{x}+\dfrac{1}{\sqrt{x}}}.$
  9. Siendo $f(x)=\log\left(\log(\operatorname{sen}x)\right),$ calcular $f’(\pi/4).$
    Solución
  1. $f’(x)=\dfrac{-1}{\sqrt{1-\left(\dfrac{1-x^2}{1+x^2}\right)^2}}\dfrac{-2x(1+x^2)-2x(1-x^2)}{(1+x^2)^2}\\
    =\dfrac{-1}{\sqrt{\dfrac{(1+x^2)^2-(1-x^2)^2}{(1+x^2)^2}}}\dfrac{-4x}{(1+x^2)^2}
    =\dfrac{4x}{\sqrt{4x^2}(1+x^2)}=\dfrac{2}{1+x^2}.$
  2. $f’(x)=\dfrac{d}{dx}\left(\dfrac{1}{2}\left(\log (\tan x+1)-\log (\tan x-1)\right)\right)$
    $=\dfrac{1}{2}\left(\dfrac{1+\tan^2 x}{\tan x+1}-\dfrac{1+\tan^2 x}{\tan x-1}\right)$
    $=\dfrac{1}{2}\dfrac{\tan x+\tan^3 x-1-\tan^2 x-\tan x-1-\tan^3 x-\tan^2 x}{\tan^2 x-1}$
    $=\dfrac{1+\tan^2 x}{1-\tan^2 x}=\dfrac{\cos^2 x+\operatorname{sen}^2 x}{\cos^2 x-\operatorname{sen}^2 x}=\dfrac{1}{\cos 2x}.$
  3. Podemos escribir $f(x)=\operatorname{senh} \left(xx^{1/2}x^{1/4}\right)=\operatorname{senh} x^{7/4},$ por tanto:
    $f’(x)=\left(\cosh x^{7/4}\right)\dfrac{7}{4}x^{3/4}=\dfrac{7}{4}\sqrt[4]{x^3}\cosh x^{7/4}=\dfrac{7}{4}\sqrt{x\sqrt{x}}\cosh \left(x\sqrt{x\sqrt{x}}\right).$
  4. $(a)$ $f’(x)=\dfrac{1}{\sqrt{1+\left(\dfrac{x^2}{a^2}\right)^2}}\dfrac{2x}{a^2}=\dfrac{1}{\sqrt{\dfrac{a^4+x^4}{a^4}}}\dfrac{2x}{a^2}$
    $=\dfrac{a^2}{\sqrt{a^2+x^4}}\dfrac{2x}{a^2}=\dfrac{2x}{\sqrt{a^2+x^4}}.$
    $(b)$ $g’(x)=\dfrac{1}{1-\left(\dfrac{2x}{x^2+1}\right)^2}\dfrac{2(x^2+1)-2x(2x)}{(x^2+1)^2}$
    $=\dfrac{1}{\dfrac{(x^2+1)^2-4x^2}{(x^2+1)^2}}\dfrac{-2x^2+2}{(x^2+1)^2}\\
    =\dfrac{2(1-x^2)}{x^4-2x^2+1}=\dfrac{2(1-x^2)}{(x^2-1)^2}=\dfrac{2(1-x^2)}{(1-x^2)^2}=\dfrac{2}{1-x^2.}$
  5. $\dfrac{d}{dx} \operatorname{arsh}x=\dfrac{d}{dx} \log \left(x + \sqrt{x^{2} + 1} \right) =\dfrac{1+\dfrac{2x}{2\sqrt{x^2+1}}}{x + \sqrt{x^{2} + 1}}\\ =\dfrac{ \sqrt{x^{2} + 1}+x}{\sqrt{x^2+1}\left(x + \sqrt{x^{2} + 1}\right)}=\dfrac{1}{\sqrt{x^2+1}}.$
    $\dfrac{d}{dx} \operatorname{arch}x=\dfrac{d}{dx}\log \left(x + \sqrt{x^{2} – 1} \right) =\dfrac{1+\dfrac{2x}{2\sqrt{x^2-1}}}{x + \sqrt{x^{2} – 1}} \\=\dfrac{ \sqrt{x^{2} – 1}+x}{\sqrt{x^2-1}\left(x + \sqrt{x^{2} – 1}\right)}=\dfrac{1}{\sqrt{x^2-1}}.$
    $\dfrac{d}{dx} \operatorname{arth}x=\dfrac{d}{dx}\left(\dfrac{1}{2}\log \dfrac{1 + x}{1 – x}\right)=\dfrac{1}{2}\dfrac{d}{dx}\left(\log (1 + x)-\log (1 – x)\right)\\=\dfrac{1}{2}\left(\dfrac{1}{1+x}+\dfrac{1}{1-x}\right)=\dfrac{1}{2}\dfrac{2}{1-x^2}=\dfrac{1}{1-x^2}.$
  6. $\dfrac{d}{dx}\left(\log \sqrt{\dfrac{1+\operatorname{sen} x}{1-\operatorname{sen} x}}\right)=\dfrac{d}{dx}\left(\dfrac{1}{2}\left(\log (1+\operatorname{sen} x)-\log (1-\operatorname{sen} x)\right)\right)\\
    =\dfrac{1}{2}\left(\dfrac{\cos x}{1+\operatorname{sen} x}+\dfrac{\cos x}{1-\operatorname{sen} x}\right)=\dfrac{1}{2}\;\dfrac{2\cos x}{1-\operatorname{sen}^2 x}=\dfrac{\cos x}{\cos^2 x}=\dfrac{1}{\cos x}=\sec x$
  7. $y’=\dfrac{1}{2}\sqrt{x^2+a}+\dfrac{x}{2}\dfrac{2x}{2\sqrt{x^2+a}}+\dfrac{a}{2}\dfrac{1}{x+\sqrt{x^2+a}}\left(1+\dfrac{2x}{2\sqrt{x^2+a}}\right)$
    $=\dfrac{\sqrt{x^2+a}}{2}+\dfrac{x^2}{2\sqrt{x^2+a}}+\dfrac{a}{2}\dfrac{1}{x+\sqrt{x^2+a}}\dfrac{\sqrt{x^2+a}+x}{\sqrt{x^2+a}}$
    $=\dfrac{\sqrt{x^2+a}}{2}+\dfrac{x^2}{2\sqrt{x^2+a}}+\dfrac{a}{2\sqrt{x^2+a}}=\dfrac{\sqrt{x^2+a}}{2}+\dfrac{x^2+a}{2\sqrt{x^2+a}}$
    $=\dfrac{\sqrt{x^2+a}}{2}+\dfrac{\sqrt{x^2+a}}{2}=\sqrt{x^2+a}.$
  8. $f’(x)=\dfrac{1}{2\sqrt{\sqrt{x}+\dfrac{1}{\sqrt{x}}}}\left(\dfrac{1}{2\sqrt{x}}+\dfrac{-\dfrac{1}{2\sqrt{x}}}{x}\right).$
    $f’(4)=\dfrac{1}{2\sqrt{2+\dfrac{1}{2}}}\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{1}{2\sqrt{\dfrac{5}{2}}}\dfrac{3}{16}=\dfrac{3\sqrt{2}}{8\sqrt{5}}=\dfrac{3\sqrt{10}}{40}.$
  9. $f’(x)=\dfrac{1}{\log\operatorname{sen}x}\;\dfrac{\cos x}{\operatorname{sen}x}\Rightarrow f’(\pi/4)=\dfrac{1}{\log \dfrac{\sqrt{2}}{2}}\;\dfrac{\dfrac{\sqrt{2}}{2}}{\dfrac{\sqrt{2}}{2}}=\dfrac{1}{\log \dfrac{\sqrt{2}}{2}}.$
Esta entrada fue publicada en Cálculo/Análisis. Guarda el enlace permanente.