Answer
$4t \tan^{-1} t+1$ and $\pi+1$
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, $=\dfrac{4yte^x}{t^2+1}+\dfrac{2e^x}{t^2+1}-\dfrac{e^t}{z}$
or, $\dfrac{dw}{dt}=4t \tan^{-1} t+1$
Using direct differentiation.
since, $w=2ye^x-\ln z$
or, $w=(2 \tan^{-1} t) (t^1+1)-t$
and $\dfrac{dw}{dt}=4t \tan^{-1} t+1$
Now, b) $\dfrac{dw}{dt}(1)=4(1) \tan^{-1} (1)+1=4(1) (\dfrac{\pi}{4})+1=\pi+1$