Réponse : Bonsoir,
i)
[tex]\displaystyle f'_{1}(x)=6x\sqrt{x}+\frac{1}{2\sqrt{x}}(3x^{2}+2)=\frac{6x \times 2\sqrt{x}\sqrt{x}+3x^{2}+2}{2\sqrt{x}}=\frac{6x \times 2x+3x^{2}+2}{2\sqrt{x}}\\=\frac{12x^{2}+3x^{2}+2}{2\sqrt{x}}=\frac{15x^{2}+2}{2\sqrt{x}}[/tex]
ii)
[tex]f'_{2}(x)=(2x+3)'4(2x+3)^{3}=2 \times 4(2x+3)^{3}=8(2x+3)^{3}[/tex]
iii)
[tex]\displaystyle f'_{3}(x)=(4x-2)'\frac{1}{2\sqrt{4x-2}}=\frac{4}{2\sqrt{4x-2}}=\frac{2}{\sqrt{4x-2}}[/tex]