What is the tangent of 9y^7+4x^3=8y+5x,(1,1) ?

The tangent of 9y^7+4x^3=8y+5x,(1,1) is y=-7/55 x+62/55

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tangent of 9y^7+4x^3=8y+5x,(1,1)

tangent of

9 y^{7} + 4 x^{3} = 8 y + 5 x, (1, 1)

y = - \frac{7}{55} x + \frac{62}{55}

Solution steps

Find the slope of

9 y^{7} + 4 x^{3} = 8 y + 5 x : \frac{d y}{d x} = \frac{5 - 12 x ^{2}}{63 y ^{6} - 8}

Compute the slope of

9 y^{7} + 4 x^{3} = 8 y + 5 x

(1, 1) : m = - \frac{7}{55}

Find the line with slope m=

- \frac{7}{55}

and passing through

(1, 1) : y = - \frac{7}{55} x + \frac{62}{55}

y = - \frac{7}{55} x + \frac{62}{55}

\frac{\partial}{\partial x} (cos^{5} (x^{7} y^{5}))

x y^{^{'}} + (1 - x) y = 0

\frac{d N}{d t} = N (1 - 0.0005 N)

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

\frac{d}{d x} (y - x n (y + x))

What is the tangent of 9y^7+4x^3=8y+5x,(1,1) ?
The tangent of 9y^7+4x^3=8y+5x,(1,1) is y=-7/55 x+62/55