What is the tangent of 7xy^3+5xy=12,\at x=1 ?

The tangent of 7xy^3+5xy=12,\at x=1 is y=-6/13 x+19/13

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tangent of 7xy^3+5xy=12,\at x=1

tangent of

7 x y^{3} + 5 x y = 12, at x = 1

y = - \frac{6}{13} x + \frac{19}{13}

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

7 x y^{3} + 5 x y = 12 : \frac{d y}{d x} = \frac{- 7 y ^{3} - 5 y}{x ( 5 + 21 y ^{2} )}

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{6}{13}

Find the line with slope m=

- \frac{6}{13}

and passing through

(1, 1) : y = - \frac{6}{13} x + \frac{19}{13}

y = - \frac{6}{13} x + \frac{19}{13}

What is the tangent of 7xy^3+5xy=12,\at x=1 ?
The tangent of 7xy^3+5xy=12,\at x=1 is y=-6/13 x+19/13