What is the tangent of (10-x)y^2=x^3,(2,1) ?

The tangent of (10-x)y^2=x^3,(2,1) is y= 13/16 x-5/8

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tangent of (10-x)y^2=x^3,(2,1)

tangent of

(10 - x) y^{2} = x^{3}, (2, 1)

y = \frac{13}{16} x - \frac{5}{8}

Solution steps

Find the slope of

(10 - x) y^{2} = x^{3} : \frac{d y}{d x} = \frac{3 x ^{2} + y ^{2}}{2 y ( 10 - x )}

Compute the slope of

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

(2, 1) : m = \frac{13}{16}

Find the line with slope m=

\frac{13}{16}

and passing through

(2, 1) : y = \frac{13}{16} x - \frac{5}{8}

y = \frac{13}{16} x - \frac{5}{8}

\int \frac{1}{16 x ^{2} + 4} d x

x + (x - 6)^{6}

\int x (x - 1) d x

\frac{d}{d x} (e^{2 + x} - 2 e^{x} + 2 e x)

x \to \frac{π}{3} lim (cot (3 x))

What is the tangent of (10-x)y^2=x^3,(2,1) ?
The tangent of (10-x)y^2=x^3,(2,1) is y= 13/16 x-5/8