To solve this problem, first, we will solve the given equation for y:
[tex]\begin{gathered} x=3\tan 2y, \\ \tan 2y=\frac{x}{3}, \\ 2y=\arctan (\frac{x}{3}), \\ y=\frac{\arctan(\frac{x}{3})}{2}=\frac{1}{2}\arctan (\frac{x}{3})\text{.} \end{gathered}[/tex]
Once we have the above equation, now we compute the derivative. To compute the derivative we will use the following properties of derivatives:
[tex]\begin{gathered} \frac{d}{dx}\arctan (x)=\frac{1}{x^2+1}, \\ \frac{dkf(x)}{dx}=k\frac{df(x)}{dx}. \end{gathered}[/tex]
Where k is a constant.
First, we use the second property above, and get that:
[tex]\frac{d\frac{\arctan(\frac{x}{3})}{2}}{dx}=\frac{d\arctan (\frac{x}{3})\times\frac{1}{2}}{dx}=\frac{1}{2}\frac{d\arctan (\frac{x}{3})}{dx}\text{.}[/tex]
Now, from the chain rule, we get:
[tex]\frac{dy}{dx}=\frac{1}{2}\frac{d\text{ arctan(}\frac{x}{3})}{dx}=\frac{1}{2}\frac{d\arctan (\frac{x}{3})}{dx}|_{\frac{x}{3}}\frac{d\frac{x}{3}}{dx}\text{.}[/tex]
Finally, computing the above derivatives (using the rule for the arctan), we get:
[tex]\frac{dy}{dx}=\frac{1}{2}\frac{\frac{1}{3}}{\frac{x^2}{9}+1}=\frac{1}{6}(\frac{1}{\frac{x^2}{9}+1})=\frac{3}{2(x^2+9)}.[/tex]
Answer:
[tex]\frac{3}{2(x^2+9)}.[/tex]
