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

The tangent of x^2-3xy=10,(1,-3) is y= 11/3 x-20/3

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

tangent of

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

y = \frac{11}{3} x - \frac{20}{3}

Solution steps

Find the slope of

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

Compute the slope of

x^{2} - 3 x y = 10

(1, - 3) : m = \frac{11}{3}

Find the line with slope m=

\frac{11}{3}

and passing through

(1, - 3) : y = \frac{11}{3} x - \frac{20}{3}

y = \frac{11}{3} x - \frac{20}{3}

y^{^{''}} - y = 5 sin (2 t y) (0)

\frac{d}{d x} (sin (x) + sin^{2} (x))

(0, 4), (6, - 4)

\frac{d y}{d x} = x y^{3} - x y

What is the tangent of x^2-3xy=10,(1,-3) ?
The tangent of x^2-3xy=10,(1,-3) is y= 11/3 x-20/3