What is the linear approximation to ln((1+𝑎𝑥)^𝑟) at 𝑥=0 ?

Answer 1

If y = ln(1+ax)^r = r ln(1+ax)
y' = ra/(1+ax)
y(0) = 0
y'(0) = ra

so, the tangent line at (0,0) is
y = rax