Answer
a. $\dfrac{16}{1+16t}$ and
b. $\dfrac{16}{49}$
Work Step by Step
a. Using chain rule:
$\dfrac{dw}{dt}=\dfrac{\partial w}{\partial x}\dfrac{dx}{dt}+\dfrac{\partial w}{\partial y}\dfrac{dy}{dt}+\dfrac{\partial w}{\partial z}\dfrac{dz}{dt}$
and $(\dfrac{-2x \sin t}{x^2+y^2+z^2})+(\dfrac{-2y \cos t}{x^2+y^2+z^2})+(\dfrac{4zt^{-1/2}}{x^2+y^2+z^2})=\dfrac{16}{1+16t}$
Now, by using direct differentiation.
We have $w=\ln (x^2+y^2+z^2)$
or, $w=\ln (cos^2 t+\sin ^2 t+16t)$
Also, $w=\ln (1+16t)$; $\dfrac{dw}{dt}=\dfrac{16}{(1+16t)}$
b. $\dfrac{dw}{dt}(3)=\dfrac{16}{(1+16t)}=\dfrac{16}{1+16(3)}=\dfrac{16}{49}$
