Find the gradient of the implicitly defined curve $y^2+2y=x^3+7x$ at all its intersection points with the line $x=1$.

Can you differentiate both sides with respect to $x.$

How can you rearrange your equation to find $\frac{dy}{dx}?$

Try to substitute $x=1$ into the original equation.

By differentiating both sides with respect to $x,$ we have $2y\frac{dy}{dx}+2\frac{dy}{dx}=3x^2+7$ (using the chain rule). Rearranging gives $\frac{dy}{dx}=\frac{3x^2+7}{2y+2}.$ Now we need to find all intersection points with the line $x=1.$ These are given by $y^2+2y-8=0,$ which has solutions $y\in\{-4,2\}.$ We then plug this into the derivative expression to get the gradient values $-\frac{5}{3}$ and $\frac{5}{3}.$