# calculation – let \$ f (x) = sqrt[2k]{(xa) ^ {2n} (xb) ^ {2m + 1}} \$ such that \$ k, m, n in mathbb {N} \$ now that is distinguishable from the \$ function f \$?

let $$f (x) = sqrt[2k]{(x-a) ^ {2n} (x-b) ^ {2m + 1}}$$ such as $$k, m, n in mathbb {N}$$ now what is differentiable function points $$f$$?

We suppose that $$h (x) = (x-a) ^ {2n} (x-b) ^ {2m + 1}$$ and $$R (x) = sqrt[2k]{x}$$ .
we know the function $$h$$ differentiable at all $$x in mathbb {R}$$ . and function $$R$$ differentiable to $$(x in mathbb {R} – 0)$$ so we do not know how is the function $$f$$ at $$h (a), h (b)$$ . so check:
$$lim_ {x to a} dfrac { sqrt[2k]{(x-a) ^ {2n} (x-b) ^ {2m + 1}}} {{x-a)} =?$$
$$lim_ {x to b} dfrac { sqrt[2k]{(x-a) ^ {2n} (x-b) ^ {2m + 1}}} {{x-b)} =?$$
How is it obtained?