What is the tangent of y=(4x)/(x^2-5),(3,3) ?

The tangent of y=(4x)/(x^2-5),(3,3) is y=-7/2 x+27/2

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tangent of y=(4x)/(x^2-5),(3,3)

tangent of

y = \frac{4 x}{x ^{2} - 5}, (3, 3)

y = - \frac{7}{2} x + \frac{27}{2}

Solution steps

Find the slope of

y = \frac{4 x}{x ^{2} - 5} : \frac{d y}{d x} = \frac{4 ( - x ^{2} - 5 )}{( x ^{2} - 5 ) ^{2}}

Compute the slope of

y = \frac{4 x}{x ^{2} - 5}

(3, 3) : m = - \frac{7}{2}

Find the line with slope m=

- \frac{7}{2}

and passing through

(3, 3) : y = - \frac{7}{2} x + \frac{27}{2}

y = - \frac{7}{2} x + \frac{27}{2}

\int 4 e^{x} - \frac{5}{x} d x

5 x - x^{2}, 2 x, 0, 3

∣ x - 8 ∣, \frac{x}{3}

3 sin (2 t)

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

What is the tangent of y=(4x)/(x^2-5),(3,3) ?
The tangent of y=(4x)/(x^2-5),(3,3) is y=-7/2 x+27/2