What is the tangent of y=(8x)/(x^2-1),(3,3) ?

The tangent of y=(8x)/(x^2-1),(3,3) is y=-5/4 x+27/4

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tangent of y=(8x)/(x^2-1),(3,3)

tangent of

y = \frac{8 x}{x ^{2} - 1}, (3, 3)

y = - \frac{5}{4} x + \frac{27}{4}

Solution steps

Find the slope of

y = \frac{8 x}{x ^{2} - 1} : \frac{d y}{d x} = \frac{8 ( - x ^{2} - 1 )}{( x ^{2} - 1 ) ^{2}}

Compute the slope of

y = \frac{8 x}{x ^{2} - 1}

(3, 3) : m = - \frac{5}{4}

Find the line with slope m=

- \frac{5}{4}

and passing through

(3, 3) : y = - \frac{5}{4} x + \frac{27}{4}

y = - \frac{5}{4} x + \frac{27}{4}

y^{^{''}} + 625 y = sec (25 x)

e^{5 x}

\int e^{- 1 + x} d x

\frac{\partial}{\partial r} (e^{r} sin (s t))

\frac{1}{( 1 - 3 x ^{2} )}, x = 0.577

What is the tangent of y=(8x)/(x^2-1),(3,3) ?
The tangent of y=(8x)/(x^2-1),(3,3) is y=-5/4 x+27/4