Question

Como derivar essa função F(x)= (2/3)^x^(3/2)

F'(x)= ???

Answer 1

$\displaystyle F(x)=\left(\frac{2}{3}\right)^{x^\frac{3}{2}}$
para facilitar o desenvolvimento consideraremos:
$\displaystyle \frac{2}{3}=\psi$
implicando que:
$\displaystyle F(x)=\psi^{x^{\psi^{-1}}}$
qualquer potência pode ser escrita na forma de ln:
$\mu^x=e^{x\ln\mu}$
desse jeito:
$\displaystyle F(x)=\psi^{x^{\frac{3}{2}}}=\exp\left(x^{\frac{3}{2}}\cdot\ln\left(\frac{2}{3}\right)\right)$
PS: $\exp(x)=e^x$
Agora com a regra da cadeia podemos derivar facilmente a função:
$\displaystyle F'(x)=\frac{dF}{dx}=\frac{d}{dx}\left(\exp\left(x^{\frac{3}{2}}\cdot\ln\left(\frac{2}{3}\right)\right)\right)\\\\u=x^{\frac{3}{2}}\cdot\ln\left(\frac{2}{3}\right)~~~~~~~~~~~~~~~~~~(1)\\\\\exp\left(x^{\frac{3}{2}}\cdot\ln\left(\frac{2}{3}\right)\right)=\exp(u)~~(2)\\\\i)~~~~\frac{dF}{dx}=\frac{dF}{du}\cdot\frac{du}{dx}\\\\ii)~~~\frac{dF}{dx}=\frac{d}{du}e^u\cdot\frac{d}{dx}x^{\frac{3}{2}}\cdot\ln\left(\frac{2}{3}\right)\\\\iii)~\frac{dF}{dx}=e^u\cdot\frac{3}{2}x^{\frac{3}{2}-1}\cdot\ln\left(\frac{2}{3}\right)\\\\iv)~~\frac{dF}{dx}=\frac{3}{2}\ln\left(\frac{2}{3}\right)x^{\frac{1}{2}}e^u\implies u=x^{\frac{3}{2}}\cdot\ln\left(\frac{2}{3}\right)\\\\v)~~\frac{dF}{dx}=\frac{3}{2}\ln\left(\frac{2}{3}\right)x^{\frac{1}{2}}\left(\frac{2}{3}\right)^{x^{\frac{3}{2}}}$
ou seja:
$\boxed{F'(x)=\frac{3}{2}\ln\left(\frac{2}{3}\right)\sqrt{x}\left(\frac{2}{3}\right)^{x^{\frac{3}{2}}}}$