What is the tangent of (12-x)y^2=x^3,(6,6) ?

The tangent of (12-x)y^2=x^3,(6,6) is y=2x-6

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tangent of (12-x)y^2=x^3,(6,6)

tangent of

(12 - x) y^{2} = x^{3}, (6, 6)

y = 2 x - 6

Solution steps

Find the slope of

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

Compute the slope of

(12 - x) y^{2} = x^{3}

(6, 6) : m = 2

Find the line with slope m=

2

and passing through

(6, 6) : y = 2 x - 6

y = 2 x - 6

\int (4 x^{3} + 2 x y^{2}) d x

x^{2} y^{^{'}} + 2 x y = ln (x)

\int \frac{x}{2 x + 3} d x

f (x) = \frac{x}{7 x ^{2} + 1}

\frac{x + 1}{x ^{3} + x - 2}

What is the tangent of (12-x)y^2=x^3,(6,6) ?
The tangent of (12-x)y^2=x^3,(6,6) is y=2x-6