Question

 [x^v Y_{v (x)}]= x^v Y_v-1(x)

Answer 1

To prove:

${d\over dx}[x^vY_{v}(x)]=x^vY_{v-1}(x)$

Taking Left Hand Side:

${d\over dx}[x^vY_{v}(x)]$

We have to assume $m=Y_u(x)\ \&\ n=x^v$

Now, use Leibnitz Theorem

${d\over dx}[x^vY_{v}(x)]=^1C_0(1!\times x^v \times Y_{v-1}(x))$

$=$ $x^vY_{v-1}(x)=$ Right Hand Side

Hence ${d\over dx}[x^vY_{v}(x)]=x^vY_{v-1}(x)$