What is the tangent of-3x^2+2xy+2y^3=125,(3,4) ?

The tangent of-3x^2+2xy+2y^3=125,(3,4) is y= 5/51 x+63/17

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tangent of-3x^2+2xy+2y^3=125,(3,4)

tangent of

- 3 x^{2} + 2 x y + 2 y^{3} = 125, (3, 4)

y = \frac{5}{51} x + \frac{63}{17}

Solution steps

Find the slope of

- 3 x^{2} + 2 x y + 2 y^{3} = 125 : \frac{d y}{d x} = \frac{3 x - y}{x + 3 y ^{2}}

Compute the slope of

- 3 x^{2} + 2 x y + 2 y^{3} = 125

(3, 4) : m = \frac{5}{51}

Find the line with slope m=

\frac{5}{51}

and passing through

(3, 4) : y = \frac{5}{51} x + \frac{63}{17}

y = \frac{5}{51} x + \frac{63}{17}

\frac{d}{d x} y = ln (10 x^{2} - 8)

x^{3} - 3 x^{2} + 2 x - 2

\frac{d}{d x} (I n x)

f (x) = x^{3} - 2 x^{2} + 1, at x = 1

\frac{d}{d x} (arcsin (e^{2 x}))

What is the tangent of-3x^2+2xy+2y^3=125,(3,4) ?
The tangent of-3x^2+2xy+2y^3=125,(3,4) is y= 5/51 x+63/17