Answer
$\mu=\frac{T^2}{c_p}\left(\frac{\partial(v / T)}{\partial T}\right)_P$
Work Step by Step
From Eq. 12-52 of the text, $$
c_p=\frac{1}{\mu}\left[T\left(\frac{\partial v}{\partial T}\right)_P-v\right]
$$ Expanding the partial derivative of $v T$ produces $$
\left(\frac{\partial v / T}{\partial T}\right)_P=\frac{1}{T}\left(\frac{\partial v}{\partial T}\right)_P-\frac{v}{T^2}
$$ When this is multiplied by $T^2$, the right-hand side becomes the same as the bracketed quantity above. Then, $$
\mu=\frac{T^2}{c_p}\left(\frac{\partial(v / T)}{\partial T}\right)_P
$$
