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

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

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

tangent of

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

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

Solution steps

Find the slope of

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

Compute the slope of

f (x) = \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}

\frac{d}{d x} (3 x^{2} - 5 x + 4)

x \to 3 + lim (x^{2} - 9)

\frac{d}{d x} (\frac{x ^{2}}{x ^{4} + 16})

\frac{d}{d x} (\frac{x - 2}{x + 3})

f (x) = e^{6}

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