## Thomas' Calculus 13th Edition

a. Using chain rule: $\dfrac{dw}{dt}=\dfrac{\partial w}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial w}{\partial z}\dfrac{dz}{dt}=(\dfrac{-2}{z})\sin t cos t +(\dfrac{2}{z})\sin t cos t+(\dfrac{x+y}{z^2 t^2})$ and, $\dfrac{(\cos^2 t+\sin^2 t)}{(\dfrac{1}{t^2})(t^2)}=1$ Now, by using direct differentiation. we have $w=(\dfrac{x}{z})+(\dfrac{y}{z})=\dfrac{\cos^2 t}{(1/t)}+\dfrac{\sin^2 t}{(1/t)}t$ From above part, we get $\dfrac{dw}{dt}=1$ b. Thus, $\dfrac{dw}{dt}(3)=1$