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# Practice Paper 2 Question 5

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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## Hints

• Hint 1
Can you differentiate both sides with respect to $$x.$$
• Hint 2
How can you rearrange your equation to find $$\frac{dy}{dx}?$$
• Hint 3
Try to substitute $$x=1$$ into the original equation.

## Solution

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 $$\frac{5}{3}.$$

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