Your derivation is basically fine, and the last step is straightforward. Define $\mu = \frac{1}{2m}$, and then
$$
Q_1' \propto \mu^{-3/2} \quad\Rightarrow\quad 
\ln Q_1' = -\frac{3}{2} \ln\mu +C 
$$
where $C$ is a constant, and so
$$
\frac{\partial}{\partial\mu} \ln Q_1' = -\frac{3}{2} \frac{1}{\mu}
= -3m
$$
which leads to $\langle K_i\rangle = \frac{3}{2}k_BT$.

This is actually a version of one of the standard ways of
deriving the equipartition of energy.
If the energy can be written $E=\kappa x^2 + E_0$,
where $x$ is one coordinate or one component of momentum,
and $E_0$ is independent of $x$,
then the classical partition function can always be factorized into an integral over $x$ and "the rest".
In that case we are just using the mathematical result
$$
\langle x^2\rangle = 
\frac{\int_{-\infty}^{\infty} dx \, x^2 \exp(-\alpha x^2)}{\int_{-\infty}^{\infty} dx \,\exp(-\alpha x^2)} = \frac{1}{2\alpha}
$$
where, here, $\alpha=\kappa/k_BT$.
This leads to $\langle \kappa x^2\rangle = \frac{1}{2}k_BT$ as expected.
The above ratio of integrals can be tackled by 
parameter differentiation
(as mentioned, for instance in *Mathematical Methods of Physics*
by J Mathews and RL Walker):

 1. Know that $\int_{-\infty}^{\infty} dx \,\exp(-\alpha x^2) = \sqrt{\pi/\alpha}$
 2. Differentiate both sides with respect to $\alpha$.

Your method is equivalent to this.