[tex]\text{Mai intai hai sa calculam derivata lui }4^{x-1}.\\
(4^{x-1})'=(4^x \circ (x-1) )'=((4^x)'\circ(x-1))\cdot (x-1)'=4^{x-1}\cdot \ln 4\\
f'(x)=x'\cdot 4^{x-1}+(4^{x-1})'\cdot x=4^{x-1}+4^{x-1}\cdot \ln 4\cdot x=\\
=\boxed{4^{x-1}(1+x\cdot \ln 4)} [/tex]