What is the tangent of y^3+xy^2-61=x+3y^2,\at x=3 ?

The tangent of y^3+xy^2-61=x+3y^2,\at x=3 is y=-5/16 x+79/16

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tangent of y^3+xy^2-61=x+3y^2,\at x=3

tangent of

y^{3} + x y^{2} - 61 = x + 3 y^{2}, at x = 3

y = - \frac{5}{16} x + \frac{79}{16}

Solution steps

Find the tangent point:

(3, 4)

Find the slope of

y^{3} + x y^{2} - 61 = x + 3 y^{2} : \frac{d y}{d x} = \frac{1 - y ^{2}}{y ( - 6 + 2 x + 3 y )}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{5}{16}

Find the line with slope m=

- \frac{5}{16}

and passing through

(3, 4) : y = - \frac{5}{16} x + \frac{79}{16}

y = - \frac{5}{16} x + \frac{79}{16}

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What is the tangent of y^3+xy^2-61=x+3y^2,\at x=3 ?
The tangent of y^3+xy^2-61=x+3y^2,\at x=3 is y=-5/16 x+79/16