Answer
a) $dc=0$
b) $d(cu)=c.du$
c) $d(u +v)=du+dv$
d) $d(uv)=u.dv+v.du$
e) $d(\frac{u}{v})=\frac{{u\times}d(u)-d(v){\times}u}{v^2}$
f) $d(x^n)=n.x^{n-1}$
Work Step by Step
a)
$dc=0$
Differentiate w.r.t. $x$
$=\frac{dc}{dx}dx$
$=0{dx}$
$dc=0$
b)
$ d(cu) =c du$
$=\frac{d}{dx}(cu)dx$
$=c\frac{d}{dx}(u)dx$
$=c\frac{du}{1}(1)$
$=c(du)$
$d(cu)=c.du$
c)
$d(u +v) =du + dv$
Taking L.H.S.
$d(u +v)=?$
$=\frac{d}{dx}(u+v)dx$
$=\frac{d}{dx}(u)dx+\frac{d}{dx}(v)dx$
$=\frac{du}{dx}dx+\frac{dv}{dx}dx$
$d(u +v)=du+dv$
d)
$d(uv) =vdu + udv$
Taking L.H.S.
$d(uv)=?$
$=\frac{d}{dx}(uv)dx$
$=u\frac{d}{dx}(v)dx+v\frac{d}{dx}(u)dx$
$=u\frac{dv}{dx}dx+v\frac{du}{dx}dx$
$=udv(1)+vdu(1)$
$d(uv)=u.dv+v.du$
e)
$d(\frac{u}{v})=\frac{v\frac{d}{dx}(u)-\frac{d}{dx}(v)(u)}{v^2}$
$=\frac{v\frac{d}{dx}(u)dx-\frac{d}{dx}(v)dx(u)}{v^2}$
$=\frac{{u\times}d(u)(1)-d(v)(1){\times}u}{v^2}$
$d(\frac{u}{v})=\frac{{u\times}d(u)-d(v){\times}u}{v^2}$
f)
$d(x^n)=n.x^{n-1}dx$
Taking L.H.S.
$d(x^n)=\frac{d}{dx}(x^n)dx$
$d(x^n)=n(x^{n-1})\frac{d}{dx}(x)$
$d(x^n)=n(x^{n-1})(1)$
$d(x^n)=n.x^{n-1}$
