Question

responder las siguientes funciones ​

Answer 1

Respuestas y pasos:

1) Aplicar la derivada:

a) $f\left(x\right)=\arcsin \left(e^x\right)$

$Derivada\:de\:\:f\left(x\right)=\arcsin \left(e^x\right)$

$\frac{d}{dx}\left(\arcsin \left(e^x\right)\right)$

$\mathrm{Aplicar\:la\:regla\:de\:la\:cadena}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$

$=\frac{d}{du}\left(\arcsin \left(u\right)\right)\frac{d}{dx}\left(e^x\right)$

$\frac{d}{du}\left(\arcsin \left(u\right)\right)=\frac{1}{\sqrt{1-u^2}}$

$\mathrm{Sustituir\:en\:la\:ecuacion}\:u=e^x$

$=\frac{1}{\sqrt{1-\left(e^x\right)^2}}\frac{d}{dx}\left(e^x\right)$

Desarrollar: $\frac{d}{dx}\left(e^x\right)$

$\mathrm{Aplicar\:la\:regla\:de\:derivacion}:\quad \frac{d}{dx}\left(e^x\right)=e^x$

$=\frac{1}{\sqrt{1-\left(e^x\right)^2}}e^x$

$\mathrm{Simplificar\:}$

$=\frac{e^x}{\sqrt{1-e^{2x}}}$

b) $\:f\left(x\right)=arc\tan \left(1+x^2\right)$

$Derivada\:de\:\:f\left(x\right)=arc\tan \left(1+x^2\right)$

$\mathrm{Aplicar\:la\:regla\:de\:la\:cadena}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$

$=\frac{d}{du}\left(\arctan \left(u\right)\right)\frac{d}{dx}\left(1+x^2\right)$

$=\frac{1}{u^2+1}\frac{d}{dx}\left(1+x^2\right)$

$\mathrm{Sustituir\:en\:la\:ecuacion}\:u=1+x^2$

$=\frac{1}{\left(1+x^2\right)^2+1}\frac{d}{dx}\left(1+x^2\right)$

$\frac{d}{dx}\left(1+x^2\right)=2x$

$=\frac{1}{\left(1+x^2\right)^2+1}\cdot \:2x$

$\mathrm{Simplificar\:}$

$=\frac{2x}{x^4+2x^2+2}$

c) $f\left(x\right)=\sqrt{arc\cos \left(x\right)}$

$Derivada\:de\:\:f\left(x\right)=\sqrt{arc\cos \left(x\right)}$

$\frac{d}{dx}\left(\sqrt{\arccos \left(x\right)}\right)$

$\mathrm{Aplicar\:la\:regla\:de\:la\:cadena}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$

$=\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(\arccos \left(x\right)\right)$

$=\frac{1}{2\sqrt{\arccos \left(x\right)}}\frac{d}{dx}\left(\arccos \left(x\right)\right)$

$\frac{d}{dx}\left(\arccos \left(x\right)\right)=-\frac{1}{\sqrt{1-x^2}}$

$=\frac{1}{2\sqrt{\arccos \left(x\right)}}\left(-\frac{1}{\sqrt{1-x^2}}\right)$

$\mathrm{Simplificar\:}$

$=-\frac{1}{2\sqrt{\arccos \left(x\right)}\sqrt{1-x^2}}$

2) Aplicar la derivada y tambien la de su respectiva función inversa:

a) $f\left(x\right)=\frac{5x}{e^x}$

$\frac{d}{dx}\left(\frac{5x}{e^x}\right)$

$\mathrm{Sacar\:la\:constante}:\quad \left(a\cdot f\right)'=a\cdot f\:'$

$=5\frac{d}{dx}\left(\frac{x}{e^x}\right)$

$\mathrm{Aplicar\:la\:regla\:del\:cociente}:\quad \left(\frac{f}{g}\right)^'=\frac{f\:'\cdot g-g'\cdot f}{g^2}$

$=5\cdot \frac{\frac{d}{dx}\left(x\right)e^x-\frac{d}{dx}\left(e^x\right)x}{\left(e^x\right)^2}$

$\frac{d}{dx}\left(x\right)=1$

$\frac{d}{dx}\left(e^x\right)=e^x$

$=5\cdot \frac{1\cdot \:e^x-e^xx}{\left(e^x\right)^2}$

$\mathrm{Simplificar\:}5$

$=\frac{5\left(1-x\right)}{e^x}$

Su inversa es $\:f^{-1}\left(x\right)=\frac{5x}{e^x}$

Entonces: $derivada\:de\:f^{-1}\left(x\right)=\frac{5x}{e^x}$

$f\left(x\right)^{-1}x=\frac{5x}{e^x}$

$\mathrm{Tratar\:}f\left(x\right)\mathrm{\:como\:}f\left(x\right)\left(x\right)$

$\mathrm{Derivar\:ambos\:lados}:\quad -\frac{x\frac{d}{dx}\left(f\left(x\right)\right)}{f\left(x\right)^2}+\frac{1}{f\left(x\right)}=\frac{5\left(1-x\right)}{e^x}$

$\mathrm{Despejar}\:\frac{d}{dx}\left(f\left(x\right)\right):\quad \frac{d}{dx}\left(f\left(x\right)\right)=-\frac{5f\left(x\right)^2\left(1-x\right)-e^xf\left(x\right)}{e^xx}$

$\frac{d}{dx}\left(f\left(x\right)\right)=-\frac{5f\left(x\right)^2\left(1-x\right)-e^xf\left(x\right)}{e^xx}$

b) $f\left(x\right)=x^3\ln \left(x\right)$

$\frac{d}{dx}\left(x^3\ln \left(x\right)\right)$

$\mathrm{Aplicar\:la\:regla\:del\:producto}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'$

$=\frac{d}{dx}\left(x^3\right)\ln \left(x\right)+\frac{d}{dx}\left(\ln \left(x\right)\right)x^3$

$\frac{d}{dx}\left(x^3\right)=3x^2$

$=3x^2\ln \left(x\right)+\frac{1}{x}x^3$

$=3x^2\ln \left(x\right)+x^2$

En su inversa:

$f\left(x\right)^{-1}x=x^3\ln \left(x\right)$

$\mathrm{Tratar\:}f\left(x\right)\mathrm{\:como\:}f\left(x\right)\left(x\right)$

$\mathrm{Derivar\:ambos\:lados}:\quad -\frac{x\frac{d}{dx}\left(f\left(x\right)\right)}{f\left(x\right)^2}+\frac{1}{f\left(x\right)}=3x^2\ln \left(x\right)+x^2$

$\mathrm{Despejar}\:\frac{d}{dx}\left(f\left(x\right)\right):\quad \frac{d}{dx}\left(f\left(x\right)\right)=-\frac{3x^2f\left(x\right)^2\ln \left(x\right)+x^2f\left(x\right)^2-f\left(x\right)}{x}$

$=-\frac{3x^2f\left(x\right)^2\ln \left(x\right)+x^2f\left(x\right)^2-f\left(x\right)}{x}$

Answer 2

Estamos en las mismas