$f(x)$ is differentiable function and $G\times{f(x)}$ satisfies the relation $$G\times{f(x)}= \lim_{h\to 0}\frac{({f(x+h)}){^m}-{({f(x)}}{^m})}{h}$$ where m is a definite natural number.
I have been asked to find the relation: $$G\times (f(x)+g(x))$$ the provided answer is: $$G\times f(x)[\frac{f(x)+{g(x)}^{ (m-1)}}{{f(x)}^{(m-1)}}]+G\times g(x)[\frac{f(x)+{g(x)}^{(m-1)}}{{g(x)}^{m-1}}]$$I have no idea about how to solve this kind of problem, I ask for your kind help regarding this problem, I will be also grateful if someone teach me the general theme of this problem or some perquisites required to solve this types of sums.
Well, $G*f(x)$ is just the derivative of $f(x)^m$
So $G*f(x) = mf(x)^{m-1}f'(x)$ call this eqn(1)
So $G*h(x) = mh(x)^{m-1}h'(x)$
So just substitute in $h(x)=f(x)+g(x)$
So
$$G*(f(x)+g(x)) $$ $$= m(f(x)+g(x))^{m-1}(f'(x)+g'(x))$$ $$= (mf'(x)+mg'(x))(f(x)+g(x))^{m-1}$$
From eqn (1), we get
$mf'(x) = (G*f(x))f(x)^{1-m}$
$mg'(x) = (G*f(x))g(x)^{1-m}$
plutting these back into our main equations
$$=[(G*f(x))f(x)^{1-m}+(G*g(x))g(x)^{1-m}](f(x)+g(x))^{m-1}$$
$$=(G*f(x))\left[\frac{f(x)+g(x)}{f(x)}\right]^{m-1}+(G*g(x))\left[\frac{f(x)+g(x)}{g(x)}\right]^{m-1}$$