Answer
a. $-2$
b. $\frac{2}{25}$
c. $-\frac{1}{2}$
d. $-7$
Work Step by Step
Given $u(1)=2, u'(1)=0, v(1)=5, v'(1)=-1$, we have:
a. $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}=(uv'+vu' )|_{x=1}=2(-1)+(5)(0)=-2$
b. $\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}=(\frac{vu'-uv'}{v^2})|_{x=1}=\frac{(5)(0)-(2)(-1)}{(5)^2}=\frac{2}{25}$
c. $\frac{d}{dx}(\frac{v}{u})=\frac{u\frac{dv}{dx}-v\frac{du}{dx}}{u^2}=(\frac{uv'-vu'}{u^2})|_{x=1}=\frac{(2)(-1)-(5)(0)}{(2)^2}=-\frac{1}{2}$
d. $\frac{d}{dx}(7v-2u)=(7v'-2u')|_{x=1}=7(-1)-2(0)=-7$
