What is the tangent of xy^3+5xy=36,(6,1) ?

The tangent of xy^3+5xy=36,(6,1) is y=-1/8 x+7/4

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tangent of xy^3+5xy=36,(6,1)

tangent of

x y^{3} + 5 x y = 36, (6, 1)

y = - \frac{1}{8} x + \frac{7}{4}

Solution steps

Find the slope of

x y^{3} + 5 x y = 36 : \frac{d y}{d x} = \frac{- y ^{3} - 5 y}{x ( 5 + 3 y ^{2} )}

Compute the slope of

x y^{3} + 5 x y = 36

(6, 1) : m = - \frac{1}{8}

Find the line with slope m=

- \frac{1}{8}

and passing through

(6, 1) : y = - \frac{1}{8} x + \frac{7}{4}

y = - \frac{1}{8} x + \frac{7}{4}

\frac{\partial}{\partial x} (3 e x sin (y))

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

\int - \frac{sin ^{2} ( x )}{cos ( x )} d x

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

\frac{sin ( 4 t )}{cos ( 6 t )}

What is the tangent of xy^3+5xy=36,(6,1) ?
The tangent of xy^3+5xy=36,(6,1) is y=-1/8 x+7/4