What is the tangent of y^3x+y^2x^2=6,(2,1) ?

The tangent of y^3x+y^2x^2=6,(2,1) is y=-5/14 x+12/7

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tangent of y^3x+y^2x^2=6,(2,1)

tangent of

y^{3} x + y^{2} x^{2} = 6, (2, 1)

y = - \frac{5}{14} x + \frac{12}{7}

Solution steps

Find the slope of

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

Compute the slope of

y^{3} x + y^{2} x^{2} = 6

(2, 1) : m = - \frac{5}{14}

Find the line with slope m=

- \frac{5}{14}

and passing through

(2, 1) : y = - \frac{5}{14} x + \frac{12}{7}

y = - \frac{5}{14} x + \frac{12}{7}

y^{^{''}} - 2 y^{^{'}} + 7 y = 0

11 x - 8 y x^{2} + 1 \frac{d y}{d x} = 0

\frac{d}{d x} (\frac{1}{10 x})

\int_{0}^{l n (2)} 10 e^{2 t} d t

\int (- \frac{1}{x} + 7 e^{x}) d x

What is the tangent of y^3x+y^2x^2=6,(2,1) ?
The tangent of y^3x+y^2x^2=6,(2,1) is y=-5/14 x+12/7