The slope of the tangent to the curve (y – x5)2 = x(1 + x2)2 at the point (1, 3) is :

Engineering

Mathematics

Tangent and Normal

Question

The slope of the tangent to the curve (y – x⁵)² = x(1 + x²)² at the point (1, 3) is :

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Differentiate each term on both sides with respect to x, we get

2(y – x²) $(\frac{dy}{dx} - 5 x^{4})$ = x · 2(1 + x²) · 2x + (1 + x²)²

Put (x = 1, y = 3), we get

${\frac{dy}{dx}]}_{(1, 3)} = 8$