How is $\frac{ds}{dt}$ related to $\frac{dx}{dt}$? The problem states: Let $x$ and $y$ be differentiable functions of $t$, and let $s = \sqrt{4x^2+6y^2}$ be a function of $x$ and $y$. How is $\frac{ds}{dt}$ related to $\frac{dx}{dt}$ if $y$ is constant?
My attempt:
\begin{align}
\frac{ds}{dt} & = \frac{d}{dt} \left( \sqrt{4x^2+6y^2} \right) \\
 & = \frac{1}{2\sqrt{4x^2+6y^2}}*\left(8x\frac{dx}{dt}+12y\frac{dy}{dt}\right)
\end{align}
If $y$ is constant, then $\frac{dy}{dt}=0$
\begin{align}
\frac{ds}{dt} & = \frac{1}{2\sqrt{4x^2+6y^2}}*\left(8x\frac{dx}{dt}+12y*0\right)\\
& =\left(\frac{8x}{2\sqrt{4x^2+6y^2}}\right)*\frac{dx}{dt}
\end{align}
This is the relation I have found, but it is not correct. What is the relation between $\frac{ds}{dt}$ and $\frac{dx}{dt}$ ?

This is an example of the usefulness of Leibniz notation in dealing with these sorts of problems. 
$$\frac{ds}{dt} = \frac{ds}{dx} \cdot \frac{dx}{dt}$$
This, intuitively, is because the $dx$'s "cancel out". 
All to do here is to find $\frac{ds}{dx}$. 
Since $y$ is not a function of $x$, we can treat it as constant. Therefore: 
$$\frac{ds}{dt}=\frac{dx}{dt} \cdot \frac{d}{dx} \left[\sqrt{4x^2+6y^2} \right]$$
And since $$\frac{d}{dx} \left[\sqrt{4x^2+6y^2} \right] = \frac{4x}{\sqrt{4x^2+6y^2}}$$
The only difference I can find is that you did not simplify your answer. To find out more, how exactly did you know your answer is incorrect?
