= Solution
Use the fixed exponents $p,q$ in the original PDF. Assume positive $a,b$ so arbitrary real powers and a relative change are defined. Set $h=k/100$. With $b$ fixed, the exact new-to-old ratio is $(1+h)^p$. The <binomial series>, or the one-variable case of the <second-order relative change of a monomial>, gives
$$
100\frac{\Delta f}{f}=100\left[ph+\frac{p(p-1)}2h^2+O(h^3)\right].
$$
Therefore the approximate percentage change through second order is
$$
\boxed{pk+\frac{p(p-1)}{200}k^2.}
$$
For negative $k$ this same signed expression describes a decrease.
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