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