Given:
[tex]8\sqrt{y}=x-5y[/tex]
Find: differentiation
Explanation: on differentitaion with respect to x
[tex]\begin{gathered} 8\sqrt{y}=x-5y \\ \frac{8}{2}y^{\frac{-1}{2}}\frac{dy}{dx}=1-5\frac{dy}{dx} \\ 4y^{\frac{-1}{2}}\frac{dy}{dx}+5\frac{dy}{dx}=1 \\ (4y^{\frac{-1}{2}}+5)\frac{dy}{dx}=1 \\ \frac{dy}{dx}=\frac{1}{4y^{\frac{-1}{2}}+5} \end{gathered}[/tex]
[tex]\begin{gathered} \frac{-4}{2}y^{\frac{-3}{2}}\frac{dy}{dx}\frac{d^2y}{dx^2}+5\frac{d^2y}{dx^2} \\ 0=(-2y^{\frac{-3}{2}}\frac{dy}{dx}+5)\frac{d^2y}{dx^2} \\ \frac{d^2y}{dx^2}=0 \end{gathered}[/tex]
put the value of
[tex]\frac{dy}{dx}[/tex]
we get,
[tex]\begin{gathered} (-2\frac{y^{\frac{-3}{2}}}{4y^{\frac{-1}{2}}+5}+5)\frac{d^2y}{dx^2}=0 \\ \frac{d^2y}{dx^2}=0 \end{gathered}[/tex]
