Question

ayudenme con las siguientes funciones es urgente ​

Answer 1

Continuación

Respuestas y Pasos:

c) $f\left(x\right)=\frac{3x^3}{e^x}$

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

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

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

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

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

$=3\cdot \frac{3x^2e^x-e^xx^3}{\left(e^x\right)^2}$

$\mathrm{Simplificar\:}$

$=\frac{3x^2\left(3-x\right)}{e^x}$

Entonces en su inversa:

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

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

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

$\mathrm{Despejar}\:\frac{d}{dx}\left(f\left(x\right)\right):$

$\frac{d}{dx}\left(f\left(x\right)\right)=-\frac{3x^2f\left(x\right)^2\left(3-x\right)-e^xf\left(x\right)}{e^xx}$

d) $f\left(x\right)=\log \left(3x\right)$

$\frac{d}{dx}\left(\log _{10}\left(3x\right)\right)$

$\mathrm{Aplicar\:las\:propiedades\:de\:los\:logaritmos}:\quad \log _a\left(b\right)=\frac{\ln \left(b\right)}{\ln \left(a\right)}$

$=\frac{d}{dx}\left(\frac{\ln \left(3x\right)}{\ln \left(10\right)}\right)$

$=\frac{1}{\ln \left(10\right)}\frac{d}{dx}\left(\ln \left(3x\right)\right)$

$\mathrm{Aplicar\:la\:regla\:de\:la\:cadena}:$

$=\frac{1}{3x}\frac{d}{dx}\left(3x\right)$

$=\frac{1}{\ln \left(10\right)x}$

En su inversa:

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

$\frac{d}{dx}\left(\frac{10^x}{3}\right)$

$=\frac{1}{3}\cdot \:10^x\ln \left(10\right)$

e) $\:f\left(x\right)=\log _3\left(x^2+1\right)$

$\frac{d}{dx}\left(\log _3\left(x^2+1\right)\right)$

$\mathrm{Aplicar\:la\:regla\:de\:la\:cadena}:\quad$

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

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

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

En su inversa:

$\frac{d}{dx}\left(\sqrt{3^x-1}\right)$

$=\frac{1}{2\sqrt{3^x-1}}\frac{d}{dx}\left(3^x-1\right)$

$=\frac{1}{2\sqrt{3^x-1}}\ln \left(3\right)\cdot \:3^x$

$=\frac{\ln \left(3\right)\cdot \:3^x}{2\sqrt{3^x-1}}$

f) $f\left(x\right)=\frac{\log _{10}\left(x\right)}{x}$

$\frac{d}{dx}\left(\frac{\log _{10}\left(x\right)}{x}\right)$

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

$=\frac{\frac{d}{dx}\left(\log _{10}\left(x\right)\right)x-\frac{d}{dx}\left(x\right)\log _{10}\left(x\right)}{x^2}$

$\frac{d}{dx}\left(\log _{10}\left(x\right)\right)=\frac{1}{x\ln \left(10\right)}$

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

$\mathrm{Simplificar\:}$

$=\frac{1-\ln \left(x\right)}{\ln \left(10\right)x^2}$

En su inversa:

$f\left(x\right)^{-1}x=\frac{\log _{10}\left(x\right)}{x}$

$\mathrm{Derivar\:ambos\:lados}:\quad -\frac{x\frac{d}{dx}\left(f\left(x\right)\right)}{f\left(x\right)^2}+\frac{1}{f\left(x\right)}=\frac{1-\ln \left(x\right)}{\ln \left(10\right)x^2}$

$=-\frac{f\left(x\right)^2\left(1-\ln \left(x\right)\right)-\ln \left(10\right)x^2f\left(x\right)}{\ln \left(10\right)x^3}$

g) $\:f\left(x\right)=\frac{e^x}{e^x+1}$

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

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

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

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

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

En su inversa:

$\frac{d}{dx}\left(\ln \left(\frac{-2x+1+\sqrt{-4x+1}}{2x}\right)\right)$

$=-\frac{1}{x\sqrt{-4x+1}}$

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

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

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

$\mathrm{Simplificar\:}\frac{e^{2x}\cdot \:2\left(x-1\right)-1\cdot \:e^{2x}}{\left(x-1\right)^2}$

$=\frac{2e^{2x}x-3e^{2x}}{\left(x-1\right)^2}$

En su inversa:

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

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

$\mathrm{Despejar}\:\frac{d}{dx}\left(f\left(x\right)\right)$

$=-\frac{f\left(x\right)^2\left(2e^{2x}x-3e^{2x}\right)-f\left(x\right)\left(x-1\right)^2}{x\left(x-1\right)^2}$