What is the tangent of xy^8+8xy=36,(4,1) ?

The tangent of xy^8+8xy=36,(4,1) is y=-9/64 x+25/16

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tangent of xy^8+8xy=36,(4,1)

tangent of

x y^{8} + 8 x y = 36, (4, 1)

y = - \frac{9}{64} x + \frac{25}{16}

Solution steps

Find the slope of

x y^{8} + 8 x y = 36 : \frac{d y}{d x} = \frac{- y ^{8} - 8 y}{8 x ( y ^{7} + 1 )}

Compute the slope of

x y^{8} + 8 x y = 36

(4, 1) : m = - \frac{9}{64}

Find the line with slope m=

- \frac{9}{64}

and passing through

(4, 1) : y = - \frac{9}{64} x + \frac{25}{16}

y = - \frac{9}{64} x + \frac{25}{16}

\frac{d}{d x} (\frac{13}{3} x^{3} + \frac{21}{2} x^{2} - 10 x - 7)

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

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

f (x) = sin (sin (x)), (3 π, 0)

f (x) = 1000 - x - \frac{1}{8} x^{2}

What is the tangent of xy^8+8xy=36,(4,1) ?
The tangent of xy^8+8xy=36,(4,1) is y=-9/64 x+25/16