Find the gradient of the implicitly defined curve
$y^2+2y=x^3+7x$ at all its intersection points with the line $x=1$.
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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.
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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$\frac{5}{3}.$