Question

Drive the following formulas;

[x^vY_v(x)]=x^vY_v-1(x)

Answer 1

Taking LHS

$\dfrac{d}{dx}[x^vY_v(x)]$

Now let $m=Y_v(x),n=x^v$

Now using Lebinitz theorem we have-

$\dfrac{d}{dx}[x^vY_v(x)]$

$=^1C_0(1! \times x^v \times Y_{v-1}(x))$

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