A particle is moving along the curve $y=3\sqrt{5x+6}$. As the particle passes through the point $(2,12)$, its $x$-coordinate increases at a rate of $3$ units per second. Find the rate of change of the distance from the particle to the origin at this instant.
I understand that $\dfrac{\mathrm dx}{\mathrm dt} = 3$, that $\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}$, and that $\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm dy}{\mathrm dt} \cdot \dfrac{1}{3}$.
I also know that the derivative of $y = \sqrt{x}$ is $\dfrac{1}{2\sqrt{x}}$. I then get $\dfrac{dy}{dt} = \dfrac{15}{2\sqrt{5x+6}} \cdot \dfrac{dx}{dt}$. What steps do I take next? Did I make any errors in reasoning? Would 45/8 be correct as the $\dfrac{dy}{dt}$ term? I tried the suggestion in the answers, but it doesn't work. What do I do next? If someone provides a full step-by-step summary of how to solve, I might even catch my own mistake.
