What is the tangent of x^3+4y^2=12,(2,-1) ?

The tangent of x^3+4y^2=12,(2,-1) is y= 3/2 x-4

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tangent of x^3+4y^2=12,(2,-1)

tangent of

x^{3} + 4 y^{2} = 12, (2, - 1)

y = \frac{3}{2} x - 4

Solution steps

Find the slope of

x^{3} + 4 y^{2} = 12 : \frac{d y}{d x} = - \frac{3 x ^{2}}{8 y}

Compute the slope of

x^{3} + 4 y^{2} = 12

(2, - 1) : m = \frac{3}{2}

Find the line with slope m=

\frac{3}{2}

and passing through

(2, - 1) : y = \frac{3}{2} x - 4

y = \frac{3}{2} x - 4

\frac{d y}{d x} = x - \frac{y}{x}

y^{^{'}} + 5 y = 4

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

f (y) = arcsin (y^{2})

\int e^{8 x} (8) d x

What is the tangent of x^3+4y^2=12,(2,-1) ?
The tangent of x^3+4y^2=12,(2,-1) is y= 3/2 x-4