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