Answer
$$5$$
Work Step by Step
The partial derivatives at $t=1$ are as follows:
$$x_t=\dfrac{1}{\sqrt t}=1;y=t-1+\ln t=0; \dfrac{dy}{dt}=1+\dfrac{1}{t}=2$$
and $$z=\pi t=\pi ;\\ \dfrac{dz}{dt}=\pi ;\\w=xe^{y}+y \ \sin \ z-\cos \ z; \\w_{x}=e^{y}=1$$
Next, $$w_{y}=xe^{y}+\sin \ z \implies w_y(t=1)=2; \\w_{z}=y \cos(z)+ \sin(z) \implies w_z(t=1)=0$$
Now, $$\dfrac{dw}{dt}=w_{x} \dfrac{dx}{dt}+w_{y}\dfrac{dy}{dt}+w_{z}\dfrac{dz}{dt}$
or, $w_t=1+4+0=5$$
