Bonjour,
Exercice 2 :
Question 1 :
[tex]f(x) = xe^x + 3x - 1\\[/tex]
u(x) = x v(x) = eˣ
u'(x) = 1 v'(x) = eˣ
[tex]f'(x) = 1 \times e^x + x \times e^x + 3 = e^x + xe^x + 3[/tex]
Question 2 :
[tex]g(x) = (x^2 + 2x -1)e^x[/tex]
[tex]u(x) = x^2 + 2x - 1[/tex] [tex]v(x) = e^x[/tex]
[tex]u'(x) = 2x + 2[/tex] [tex]v'(x) = e^x[/tex]
[tex]g'(x) = (2x + 2) \times e^x + (x^2 + 2x - 1) \times e^x = 2xe^x + 2e^x + x^2e^x + 2xe^x - e^x[/tex]
[tex]= x^2e^x + 4xe^x + e^x[/tex]
[tex]= (x^2 + 4x + 1)e^x[/tex]
Question 3 :
[tex]h(x) = \frac{e^x}{e^x + x}[/tex]
[tex]u(x) = e^x[/tex] [tex]v(x) = e^x + x[/tex]
[tex]u'(x) = e^x[/tex] [tex]v'(x) = e^x + 1[/tex]
[tex]h'(x) = \frac{u'v - uv'}{v^2} = \frac{e^x \times (e^x + x) - e^x \times (e^x + 1)}{(e^x + x)^2} = \frac{e^{2x} + xe^x -e^{2x} -e^x}{(e^x + x)^2} = \frac{ xe^x -e^x}{(e^x + x)^2} = \frac{ (x - 1)e^x}{(e^x + x)^2}[/tex]
