What is the tangent of y= 2/(x-2)(4.1) ?

The tangent of y= 2/(x-2)(4.1) is y=-(8.2)/((a_{0)-2)^2}x+(16.4a_{0}-16.4)/((a_{0)-2)^2}

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tangent of y= 2/(x-2)(4.1)

tangent of

y = \frac{2}{x - 2} (4.1)

y = - \frac{8.2}{( a _{0} - 2 ) ^{2}} x + \frac{16.4 a _{0} - 16.4}{( a _{0} - 2 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{8.2}{a _{0} - 2})

Find the slope of

y = \frac{2}{x - 2} 4.1 : \frac{d y}{d x} = - \frac{8.2}{( x - 2 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{8.2}{( a _{0} - 2 ) ^{2}}

Find the line with slope m=

- \frac{8.2}{( a _{0} - 2 ) ^{2}}

and passing through

(a_{0}, \frac{8.2}{a _{0} - 2}) : y = - \frac{8.2}{( a _{0} - 2 ) ^{2}} x + \frac{16.4 a _{0} - 16.4}{( a _{0} - 2 ) ^{2}}

y = - \frac{8.2}{( a _{0} - 2 ) ^{2}} x + \frac{16.4 a _{0} - 16.4}{( a _{0} - 2 ) ^{2}}

x \to 0 + lim ((2 x)^{5 x})

\int \frac{1}{y ^{2}} d y

x \to 3 lim (\frac{8}{x - 3})

- x^{2} + 4 (x)

\int x (sec^{2} (x) + 2 tan^{2} (x)) d x

What is the tangent of y= 2/(x-2)(4.1) ?
The tangent of y= 2/(x-2)(4.1) is y=-(8.2)/((a_{0)-2)^2}x+(16.4a_{0}-16.4)/((a_{0)-2)^2}