## Calculus 8th Edition

$$\frac{\partial z}{\partial s}=\frac{2(s\ln^2t+t^2e^{2s})}{1+(s^2\ln^2t+t^2e^{2s})^2}$$ $$\frac{\partial z}{\partial t}=\frac{2s^2\ln t+2t^2e^{2s}}{t(1+(s^2\ln^2t+t^2e^{2s})^2)}$$
The partial derivative with respect to $s$ is: $$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}= \frac{\partial}{\partial x}(\arctan(x^2+y^2))\frac{\partial }{\partial s}(s\ln t)+\frac{\partial}{\partial y}(\arctan (x^2+y^2))\frac{\partial}{\partial s}(te^s)= \frac{1}{1+(x^2+y^2)^2}\frac{\partial}{\partial x}(x^2+y^2)\cdot\ln t+ \frac{1}{1+(x^2+y^2)^2}\frac{\partial }{\partial y}(x^2+y^2)\cdot te^s= \frac{1}{1+(x^2+y^2)^2}\cdot2x\cdot \ln t+\frac{1}{1+(x^2+y^2)^2}\cdot 2y\cdot te^s= \frac{2x\ln t+2yte^s}{1+(x^2+y^2)^2}$$ Now we will express solution in terms of $s$ and $t$: $$\frac{\partial z}{\partial s}=\frac{2x\ln t+2yte^s}{1+(x^2+y^2)^2}=\frac{2s\ln t\cdot \ln t+2te^s\cdot te^s}{1+(s^2\ln^2t+t^2e^{2s})^2}= \frac{2(s\ln^2t+t^2e^{2s})}{1+(s^2\ln^2t+t^2e^{2s})^2}$$ The partial derivative with respect to $t$ is: $$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}= \frac{\partial}{\partial x}(\arctan(x^2+y^2))\frac{\partial }{\partial t}(s\ln t)+\frac{\partial}{\partial y}(\arctan(x^2+y^2))\frac{\partial}{\partial t}(te^s)= \frac{1}{1+(x^2+y^2)^2}\frac{\partial}{\partial x}(x^2+y^2)\cdot\frac{s}{t}+\frac{1}{1+(x^2+y^2)^2}\frac{\partial}{\partial y}(x^2+y^2)\cdot e^s= \frac{1}{1+(x^2+y^2)^2}\cdot2x\frac{s}{t}+\frac{1}{1+(x^2+y^2)^2}\cdot2y\cdot e^s= \frac{2xs+2ye^st}{t(1+(x^2+y^2)^2)}$$ Now we will express solution in terms pf $s$ and $t$: $$\frac{\partial z}{\partial t}=\frac{2xs+2ye^st}{t(1+(x^2+y^2)^2)}= \frac{2s\ln t\cdot s+2te^s\cdot e^st}{t(1+(s^2\ln^2t+t^2e^{2s})^2)}= \frac{2s^2\ln t+2t^2e^{2s}}{t(1+(s^2\ln^2t+t^2e^{2s})^2)}$$