Answer
${\dfrac{ds}{dt}}={\dfrac{x}{\sqrt{(x^2+y^2)}}({\dfrac{dx}{dt}})}$
${\dfrac{ds}{dt}}={\dfrac{1}{\sqrt{(x^2+y^2)}}(x{\dfrac{dx}{dt}+y{\frac{dy}{dt}}})}$
and
${\dfrac{dx}{dt}}={\dfrac{-y}{x}\dfrac{dy}{dt}}$
Work Step by Step
Given: $s={\sqrt{(x^2+y^2)}}$a)
Now, take $y$ as constant and differentiating $s$, we get:
Thus, ${\dfrac{ds}{dt}}={\frac{1}{2}(x^2+y^2)^{\frac{-1}{2}}2x{\frac{dx}{dt}}}$
${\dfrac{ds}{dt}}={\frac{x}{\sqrt{(x^2+y^2)}}({\frac{dx}{dt}})}$
Now, differentiating $s$ for both variables $x,y$, we have
${\dfrac{ds}{dt}}={\frac{1}{2}(x^2+y^2)^{\frac{-1}{2}}(2x{\frac{dx}{dt}}+2y{\frac{dy}{dt}})}={\frac{1}{\sqrt{(x^2+y^2)}}(x{\frac{dx}{dt}+y{\frac{dy}{dt}}})}$
or, $0={\dfrac{1}{\sqrt{(x^2+y^2)}}(x{\frac{dx}{dt}+y{\frac{dy}{dt}}})}$
or, ${\dfrac{dx}{dt}}={\dfrac{-y}{x}\frac{dy}{dt}}$
Hence,
${\dfrac{ds}{dt}}={\dfrac{x}{\sqrt{(x^2+y^2)}}({\dfrac{dx}{dt}})}$
${\dfrac{ds}{dt}}={\dfrac{1}{\sqrt{(x^2+y^2)}}(x{\dfrac{dx}{dt}+y{\frac{dy}{dt}}})}$
and
${\dfrac{dx}{dt}}={\dfrac{-y}{x}\dfrac{dy}{dt}}$