Question

Drive the following formulas;

d/dx [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 the Leibnitz theorem

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

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

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