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

The tangent of (4-x)y^2=x^3,(2,2) is y=2x-2

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

tangent of

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

y = 2 x - 2

Solution steps

Find the slope of

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

Compute the slope of

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

(2, 2) : m = 2

Find the line with slope m=

2

and passing through

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

y = 2 x - 2

\frac{\partial}{\partial y} (- \frac{y}{x ^{2}})

\int 4 x^{7} d x

f (x) = 0.0622 x^{2} + 1.1119 x + 2.3979

y = x^{5}

\int \frac{10}{x} d x

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