If \( z=\ln \left(x^{2}+y^{2}\right) \), where \( \mathrm{x}=e^{t} \) and \( \mathrm{y}=t^{2} \), then \( \frac{d z}{d t}= \)

To find \(\frac{dz}{dt}\) for the function \( z = \ln(x^2 + y^2) \) where \( x = e^t \) and \( y = t^2 \), follow these steps: 1. **Differentiate \( z \) with respect to \( t \):** \[ \frac{dz}{dt} = \frac{d}{dt} \left( \ln(x^2 + y^2) \right) = \frac{1}{x^2 + y^2} \cdot \frac{d}{dt}(x^2 + y^2) \] 2. **Compute the derivatives of \( x^2 \) and \( y^2 \) with respect to \( t \):** - For \( x = e^t \): \[ \frac{dx}{dt} = e^t \quad \Rightarrow \quad \frac{d}{dt}(x^2) = 2x \cdot \frac{dx}{dt} = 2e^{2t} \] - For \( y = t^2 \): \[ \frac{dy}{dt} = 2t \quad \Rightarrow \quad \frac{d}{dt}(y^2) = 2y \cdot \frac{dy}{dt} = 4t^3 \] 3. **Combine the results:** \[ \frac{d}{dt}(x^2 + y^2) = 2e^{2t} + 4t^3 \] 4. **Substitute back into the expression for \(\frac{dz}{dt}\):** \[ \frac{dz}{dt} = \frac{2e^{2t} + 4t^3}{e^{2t} + t^4} \] **Final Answer:** \[ \frac{dz}{dt} = \frac{\,2\,e^{2t} + 4\,t^{3}\,}{\,e^{2t} + t^{4}\,} \]