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?