Answer
$$y'=\frac{3}{2t}+\frac{1}{2\sqrt t}$$
Work Step by Step
$$y=\ln(t^{3/2})+\sqrt t=\ln(t^{3/2})+t^{1/2}$$
Recall the Derivative Rule for Natural Logarithm: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$
Therefore, we have $$y'=(\ln(t^{3/2})+t^{1/2})'=\frac{1}{t^{3/2}}(t^{3/2})'+\frac{1}{2}t^{-1/2}$$ $$y'=\frac{1}{t^{3/2}}\frac{3}{2}t^{1/2}+\frac{1}{2t^{1/2}}=\frac{3t^{1/2}}{2t^{3/2}}+\frac{1}{2\sqrt t}$$ $$y'=\frac{3}{2t}+\frac{1}{2\sqrt t}$$