## Calculus 8th Edition

$$\frac{\partial}{\partial t}p=\frac{2t^3}{\sqrt{t^4+u^2\cos v}};$$ $$\frac{\partial}{\partial v}p=-\frac{u^2\sin v}{2\sqrt{t^4+u^2\cos v}};$$ $$\frac{\partial}{\partial u}p=\frac{u\cos v}{\sqrt{t^4+u^2\cos v}}.$$
Regarding $u$ and $v$ as constant and differentiating with respect to $t$ we get $$\frac{\partial}{\partial t}p=\frac{\partial}{\partial t}\sqrt{t^4+u^2\cos v}=\frac{1}{2\sqrt{t^4+u^2\cos v}}\frac{\partial}{\partial t}(t^4+u^2\cos v)=\frac{1}{2\sqrt{t^4+u^2\cos v}}\cdot 4t^3=\frac{2t^3}{\sqrt{t^4+u^2\cos v}}.$$ Regard $u$ and $t$ as constant and differentiate with respect to $v$: $$\frac{\partial}{\partial v}p=\frac{\partial}{\partial v}\sqrt{t^4+u^2\cos v}=\frac{1}{2\sqrt{t^4+u^2\cos v}}\frac{\partial}{\partial v}(t^4+u^2\cos v)=\frac{1}{2\sqrt{t^4+u^2\cos v}}u^2(-\sin v)=-\frac{u^2\sin v}{2\sqrt{t^4+u^2\cos v}}.$$ Regard $v$ and $t$ as constant and differentiate with respect to $u$: $$\frac{\partial}{\partial u}p=\frac{\partial}{\partial u}\sqrt{t^4+u^2\cos v}=\frac{1}{2\sqrt{t^4+u^2\cos v}}\frac{\partial}{\partial u}(t^4+u^2\cos v)=\frac{1}{2\sqrt{t^4+u^2\cos v}}\cos v\cdot2u=\frac{u\cos v}{\sqrt{t^4+u^2\cos v}}.$$