So I was given the following prompt when studying for a test:
"A curve $C$ is defined by the parametric equations $x(t)=3+t^2$ and $y(t)=t^3+5t$. Find an equation of the line tangent to the graph of $C$ at the point where $t=1$."
I know this is a bit similar to a question that was asked a while ago, but I was able to work through a bit more of the problem. I ended up reducing the problem down to $y-6=\frac{dy}{dx}(x-4)$, but I'm a bit confused about what $\frac{dy}{dx}$ would actually look like here. I ended up finding an equation for it, but I thought that it had to be a number in the context of this equation? (The equation I found was as follows: $\frac{3t^2+5}{2t}$) Any help would be appreciated!
The question wants the tangent line where $t = 1$, so you should plug in $t = 1$ into your (correct) formula for $dy/dx$.