Question 12.1.2: Find dz/dt when z = f (x, y) = xe^2y with x = √t and y = ln ......

Here f_{1}^{\prime}(x,y)=e^{2y},f_{2}^{\prime}(x,y)=2x e^{2y},\;\mathrm{d}x/\mathrm{d}t=1/2\sqrt{t},\;\mathrm{and}\;\mathrm{d}y/\mathrm{d}t=1/t. Now y = ln t implies that e^{2 y}=e^{2\ln t}=(e^{\ln t})^{2}=t^{2}, so formula (12.1.1) gives

{\frac{\mathrm{d}z}{\mathrm{d}t}}=f_{1}^{\prime}(x,y){\frac{\mathrm{d}x}{\mathrm{d}t}}+f_{2}^{\prime}(x,y){\frac{\mathrm{d}y}{\mathrm{d}t}}       (12.1.1)

As in Example 12.1.1, we can verify the chain rule directly by  substituting x =\sqrt{t} and y = ln t in the formula for f (x, y), implying that z = xe^{2y} =\sqrt{t }· t^{2} = t^{5/2}, whose derivative is dz/dt = \frac{5}{2}t^{3/2}.