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Point Form of Tangent: Hyperbola

The slope of ...

Question

The slope of the tangent to the curve $(y - x^{5})^{2} = x (1 + x^{2})^{2}$ at the point $(1, 3)$ is

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Solution

$(y - x^{5})^{2} = x (1 + x^{2})^{2} \Rightarrow 2 (y - x^{5}) (y^{'} - 5 x^{4}) = (1 + x^{2})^{2} + 2 x (1 + x^{2}) 2 x$
Putting $(1, 3)$
$\Rightarrow 2 (3 - 1) (y^{'} - 5) = (1 + 1)^{2} + 4 (1 + 1) \Rightarrow 4 (y^{'} - 5) = 12 \Rightarrow y^{'} = 8$

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