## Thomas' Calculus 13th Edition

$e^{t-1}-(1+\ln t) \cos (t \ln t)$ and $0$
a. Using chain rule. $\dfrac{dw}{dt}=\dfrac{\partial w}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial w}{\partial z}\dfrac{dz}{dt}$ and, $(-y) \cos (xy)-\dfrac{x \cos xy}{t}+e^{(t-1)}=e^{(t-1)}-(1+\ln t) \cos (t \ln t)$ Now, by using direct differentiation. we have, $w=2-\sin xy$ and, $w=e^{(t-1)}- \sin (t \ln t)$ and $\dfrac{dw}{dt}=e^{t-1}-(1+\ln t) \cos (t \ln t)$ b. $\dfrac{dw}{dt}(1)=e^{(t-1)}-(1+\ln t) \cos (t \ln t)=e^{(1-1)}-(1+\ln 1) \cos ( \ln 1)=1-1=0$