What is the tangent of (2x)/((x^2-5)^3),\at x=0 ?

The tangent of (2x)/((x^2-5)^3),\at x=0 is y=-2/125 x

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tangent of (2x)/((x^2-5)^3),\at x=0

tangent of

\frac{2 x}{( x ^{2} - 5 ) ^{3}}, at x = 0

y = - \frac{2}{125} x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

y = \frac{2 x}{( x ^{2} - 5 ) ^{3}} : \frac{d y}{d x} = - \frac{10 ( x ^{2} + 1 )}{( x ^{2} - 5 ) ^{4}}

Compute the slope of

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

x = 0 : m = - \frac{2}{125}

Find the line with slope m=

- \frac{2}{125}

and passing through

(0, 0) : y = - \frac{2}{125} x

y = - \frac{2}{125} x

\int \frac{1}{x ^{3} x ^{2} - 81} d x

\frac{d}{d x} ((3 x - 1) (x - 2)^{4})

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

y = 2 x sin (x)

What is the tangent of (2x)/((x^2-5)^3),\at x=0 ?
The tangent of (2x)/((x^2-5)^3),\at x=0 is y=-2/125 x