Answer with Step-by-step explanation:
We are given that u and v are functions of x and both are differentiable at x=0
[tex]u(0)=4,u'(0)=7,v(0)=2,v'(0)=1[/tex]
a.We have to find the values of [tex]\frac{d(uv)}{dx}[/tex]
[tex]\frac{d(u\cdot v)}{dx}=u'v+uv'[/tex]
Using this formula
Then , we get
[tex][\frac{d(uv)}{dx}]_{x=0}=u'(0)v(0)+u(0)v'(0)=7(2)+4(1)=14+4=18[/tex]
[tex][\frac{d(uv)}{dx}]_{x=0}=18[/tex]
b.[tex]\frac{d(u/v)}{dx}=\frac{u'v-uv'}{v^2}[/tex]
[tex][\frac{d(u/v)}{dx}]_{x=0}=\frac{u'(0)v(0)-u(0)v'(0)}{v^2(0)}=\frac{7(2)-4(1)}{2^2}=\frac{14-4}{4}=\frac{10}{4}=\frac{5}{2}[/tex]
[tex][\frac{d(u/v)}{dx}]_{x=0}=\frac{5}{2}[/tex]
c.
[tex][\frac{d(v/u)}{dx}]_{x=0}=\frac{v'(0)u(0)-v(0)u'(0)}{u^2(0)}=\frac{1(4)-7(2)}{4^2}[/tex]
[tex][\frac{d(v/u)}{dx}]_{x=0}=\frac{-10}{16}=\frac{-5}{8}[/tex]
d.[tex]\frac{d(-6v-9u)}{dx}=-6v'-9u'[/tex]
[tex][\frac{d(-6v-9u)}{dx}]_{x=0}=-6v'(0)-9u'(0)=-6(1)-9(7)=-6-63=-69[/tex]
[tex][\frac{d(-6v-9u)}{dx}]_{x=0}=-69[/tex]