[tex]\text{Se foloseste formula:}(f^g)'=f^g(f'\cdot \dfrac{g}{f}+g' \cdot \ln f),\\
Avem: f(x)=e^x,g(x)=2x+3\\
Deci: (f^g)'=e^{x(2x+3)}(e^x\cdot \dfrac{2x+3}{e^x}+2\cdot x)=e^{x(2x+3)}\cdot(4x+3)\\
Asadar:\boxed{(e^{x(2x+3)})'=e^{x(2x+3)}\cdot(4x+3)}[/tex]
