What is the tangent of 3xy^3+4xy=28,(4,1) ?

The tangent of 3xy^3+4xy=28,(4,1) is y=-7/52 x+20/13

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tangent of 3xy^3+4xy=28,(4,1)

tangent of

3 x y^{3} + 4 x y = 28, (4, 1)

y = - \frac{7}{52} x + \frac{20}{13}

Solution steps

Find the slope of

3 x y^{3} + 4 x y = 28 : \frac{d y}{d x} = \frac{- 3 y ^{3} - 4 y}{x ( 4 + 9 y ^{2} )}

Compute the slope of

3 x y^{3} + 4 x y = 28

(4, 1) : m = - \frac{7}{52}

Find the line with slope m=

- \frac{7}{52}

and passing through

(4, 1) : y = - \frac{7}{52} x + \frac{20}{13}

y = - \frac{7}{52} x + \frac{20}{13}

x \to 2 lim (3 x^{2} - x - 10)

\frac{d ^{2} y}{d x ^{2}} - 2 x \frac{d y}{d x} + 2 y = 0

y = x sin (\frac{2}{x})

x^{7} f (x)

y = x + x

What is the tangent of 3xy^3+4xy=28,(4,1) ?
The tangent of 3xy^3+4xy=28,(4,1) is y=-7/52 x+20/13