bonsoir pouvez vous m'aider svp calculer la dérivée de [tex] \frac{1 + \sin(x) }{1 + \cos(x) } [/tex] [tex] \frac{ 1 + \sin(x) }{1 - \cos(x) } [/tex] [tex] \fra

f(x)=(1+sin(x))/(1+cos(x))

f'(x)=(sin²(x)+sin(x)+cos²(x)+cos(x))/(1+cos(x))²

g(x)=(1+sin(x))/(1-cos(x))

g'(x)=(-sin(x).(1+sin(x))-cos²(x)+cos(x))/(1-cos(x))²

h(x)=(1-sin(x))/(1-cos(x))

h'(x)=(sin(x).(sin(x)-1)+cos²(x)-cos(x))/(1-cos(x))²

k(x)=5.exp(-3x)-3.exp(-5x)

k'(x)=-15.exp(-3x)+15.exp(-5x)