Find $d y / d x$
\[
\begin{array}{l}
\quad x=\frac{t}{3+t}, \quad y=\sqrt{3+t} \\
\frac{d y}{d x}=1
\end{array}
\]
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Step 1 ：Given the expressions for x and y in terms of a third variable t, we have x = \(\frac{t}{3+t}\) and y = \(\sqrt{3+t}\).

Step 2 ：We are asked to find the derivative of y with respect to x, which can be found by first finding dy/dt and dx/dt, and then dividing dy/dt by dx/dt.

Step 3 ：First, we find dx/dt. The derivative of x with respect to t, dx/dt, is \(-\frac{t}{(t + 3)^2} + \frac{1}{t + 3}\).

Step 4 ：Next, we find dy/dt. The derivative of y with respect to t, dy/dt, is \(\frac{1}{2\sqrt{t + 3}}\).

Step 5 ：Finally, we find dy/dx by dividing dy/dt by dx/dt. This gives us \(\frac{1}{2\sqrt{t + 3}\left(-\frac{t}{(t + 3)^2} + \frac{1}{t + 3}\right)}\).

Step 6 ：Thus, the derivative of y with respect to x, \(\frac{d y}{d x}\), is \(\boxed{\frac{1}{2\sqrt{t + 3}\left(-\frac{t}{(t + 3)^2} + \frac{1}{t + 3}\right)}}\).