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The slope of the tangent to the curve (y...

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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The correct Answer is:
`(8)`
Slope of tangent at the point `(x _1, y _1)` is ` ((dy)/(dx))_("("x_1, y_1")")`
Given curve, `(y - x ^(5))^(2) = x ( 1+ x ^(2))^(2)`
`rArr 2 ( y - x ^(5) ) ((dy ) /(dx) - 5x ^(4)) = (1 + x ^(2)) ^(2) + 2 x (1 + x ^(2) ) * 2x `
Put `x = 1 and y = 3, dy //dx = 8`
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