What is the tangent of x^2y^2=16,(1,4) ?

The tangent of x^2y^2=16,(1,4) is y=-4x+8

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tangent of x^2y^2=16,(1,4)

tangent of

x^{2} y^{2} = 16, (1, 4)

y = - 4 x + 8

Solution steps

Find the slope of

x^{2} y^{2} = 16 : \frac{d y}{d x} = - \frac{y}{x}

Compute the slope of

x^{2} y^{2} = 16

(1, 4) : m = - 4

Find the line with slope m=

- 4

and passing through

(1, 4) : y = - 4 x + 8

y = - 4 x + 8

\frac{\partial}{\partial x} (5 - 3 x^{2} - y^{2})

\frac{d y}{d x} - x^{2} y = 0

\int x^{7} [0.7] d x

1 - x

\frac{d}{d x} (sec (x) \cdot tan (x))

What is the tangent of x^2y^2=16,(1,4) ?
The tangent of x^2y^2=16,(1,4) is y=-4x+8