Répondre :
a)u/v => (u'v-uv')/v²
f(x)=(2x+1)/(x-3) => f'(x)=((2)(x-3)-(2x+1)(1))/(x-3)²=(2x-6-2x-1)/(x-3)²=-7/(x-3)²
b)u*v => u'v+uv'
f(x)=(x+4)(1-3x) => f'(x)=(1)(1-3x)+(x+4)(-3)=1-3x-3x-12=-6x-11
c)1/v => -v'/v²
f(x)=1/(2x+8) => f'(x)=-2/(2x+8)²
d)k*u => k*u'
f(x)=(3x-1)/5=(1/5)(3x-1) => f'(x)=(1/5)(3)=3/5