#### Answer

$\frac{ds}{dt} = \frac{x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}}{\sqrt{(x^2+y^2+z^2)}}$
$\frac{ds}{dt} = \frac{(y\frac{dy}{dt}+z\frac{dz}{dt})}{\sqrt{(x^2+y^2+z^2)}}$
$x{\frac{dx}{dt}=-(y\frac{dy}{dt}+z\frac{dz}{dt}})$

#### Work Step by Step

Given $s={\sqrt{(x^2+y^2+z^2)}}$
on differentiating both sides:
$\frac{ds^2}{dt}=\frac{d{{(x^2+y^2+z^2)}}}{dt}$
$2s\frac{ds}{dt}=2x{\frac{dx}{dt}+2y\frac{dy}{dt}+2z\frac{dz}{dt}}$
$\frac{ds}{dt} = \frac{x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}}{\sqrt{(x^2+y^2+z^2)}}$
taking x as a constant, the above equation becomes:
$s\frac{ds}{dt}=y\frac{dy}{dt}+z\frac{dz}{dt}$
$\frac{ds}{dt} = \frac{(y\frac{dy}{dt}+z\frac{dz}{dt})}{\sqrt{(x^2+y^2+z^2)}}$
on taking s as a constant:
$0=x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}$
$x{\frac{dx}{dt}=-(y\frac{dy}{dt}+z\frac{dz}{dt}})$
Thus the final answer is:
$\frac{ds}{dt} = \frac{x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}}{\sqrt{(x^2+y^2+z^2)}}$
$\frac{ds}{dt} = \frac{(y\frac{dy}{dt}+z\frac{dz}{dt})}{\sqrt{(x^2+y^2+z^2)}}$
$x{\frac{dx}{dt}=-(y\frac{dy}{dt}+z\frac{dz}{dt}})$