Partial Derivatives 2E97D6

1. **State the problem:** We are given a function $$P(x,y) = x^2 y^2 + 3x y^4$$ where $$x = f(r,t) = r^2 + e^r$$ and $$y = f(r,t) = 2t^2 r + \ln t$$. We need to find the partial derivatives $$\frac{\partial P}{\partial r}$$ and $$\frac{\partial P}{\partial t}$$. 2. **Recall the chain rule for multivariable functions:** Since $$P$$ depends on $$x$$ and $$y$$, which in turn depend on $$r$$ and $$t$$, we use the chain rule: $$\frac{\partial P}{\partial r} = \frac{\partial P}{\partial x} \cdot \frac{\partial x}{\partial r} + \frac{\partial P}{\partial y} \cdot \frac{\partial y}{\partial r}$$ $$\frac{\partial P}{\partial t} = \frac{\partial P}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial P}{\partial y} \cdot \frac{\partial y}{\partial t}$$ 3. **Compute partial derivatives of $$P$$ with respect to $$x$$ and $$y$$:** $$\frac{\partial P}{\partial x} = 2x y^2 + 3 y^4$$ $$\frac{\partial P}{\partial y} = 2 x^2 y + 12 x y^3$$ 4. **Compute partial derivatives of $$x$$ and $$y$$ with respect to $$r$$ and $$t$$:** - For $$x = r^2 + e^r$$: $$\frac{\partial x}{\partial r} = 2r + e^r$$ $$\frac{\partial x}{\partial t} = 0$$ (since $$x$$ does not depend on $$t$$) - For $$y = 2 t^2 r + \ln t$$: $$\frac{\partial y}{\partial r} = 2 t^2$$ $$\frac{\partial y}{\partial t} = 4 t r + \frac{1}{t}$$ 5. **Substitute all into the chain rule formulas:** $$\frac{\partial P}{\partial r} = (2 x y^2 + 3 y^4)(2 r + e^r) + (2 x^2 y + 12 x y^3)(2 t^2)$$ $$\frac{\partial P}{\partial t} = (2 x y^2 + 3 y^4)(0) + (2 x^2 y + 12 x y^3)(4 t r + \frac{1}{t})$$ 6. **Final answers:** $$\boxed{\frac{\partial P}{\partial r} = (2 x y^2 + 3 y^4)(2 r + e^r) + (2 x^2 y + 12 x y^3)(2 t^2)}$$ $$\boxed{\frac{\partial P}{\partial t} = (2 x^2 y + 12 x y^3) \left(4 t r + \frac{1}{t}\right)}$$ where $$x = r^2 + e^r$$ and $$y = 2 t^2 r + \ln t$$.