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

The tangent of y=(((x+2))/((x-2)))^2,\at x=3 is y=-40x+145

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tangent of y=(((x+2))/((x-2)))^2,\at x=3

tangent of

y = (\frac{( x + 2 )}{( x - 2 )})^{2}, at x = 3

y = - 40 x + 145

Solution steps

Find the tangent point:

(3, 25)

Find the slope of

y = (\frac{x + 2}{x - 2})^{2} : \frac{d y}{d x} = - \frac{8 ( x + 2 )}{( x - 2 ) ^{3}}

Compute the slope of

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

x = 3 : m = - 40

Find the line with slope m=

- 40

and passing through

(3, 25) : y = - 40 x + 145

y = - 40 x + 145

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\int \frac{x - 1}{( x + 1 ) ^{3}} d x

x \to 5 lim (x + 3)

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

What is the tangent of y=(((x+2))/((x-2)))^2,\at x=3 ?
The tangent of y=(((x+2))/((x-2)))^2,\at x=3 is y=-40x+145