What is the tangent of xy^3+xy=2,(1,1) ?

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

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tangent of xy^3+xy=2,(1,1)

tangent of

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

y = - \frac{1}{2} x + \frac{3}{2}

Solution steps

Find the slope of

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

Compute the slope of

x y^{3} + x y = 2

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

Find the line with slope m=

- \frac{1}{2}

and passing through

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

y = - \frac{1}{2} x + \frac{3}{2}

f (x) = \frac{4}{x ^{2}}, at x = 1

\frac{d}{d x} (\frac{5 x ^{3} + 7 x ^{2}}{x})

\frac{d}{d x} (x - \frac{1}{9} x)

x \to 0 lim (e)

n = 1 \sum \infty \frac{n - 1}{3 n - 1}

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