What is the tangent of 2/(x-8)(10.1) ?

The tangent of 2/(x-8)(10.1) is y=-(20.2)/((a_{0)-8)^2}x+(40.4a_{0}-161.6)/((a_{0)-8)^2}

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tangent of 2/(x-8)(10.1)

tangent of

\frac{2}{x - 8} (10.1)

y = - \frac{20.2}{( a _{0} - 8 ) ^{2}} x + \frac{40.4 a _{0} - 161.6}{( a _{0} - 8 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{20.2}{a _{0} - 8})

Find the slope of

y = \frac{2}{x - 8} 10.1 : \frac{d y}{d x} = - \frac{20.2}{( x - 8 ) ^{2}}

Compute the slope of

y = \frac{2}{x - 8} 10.1

x = a_{0} : m = - \frac{20.2}{( a _{0} - 8 ) ^{2}}

Find the line with slope m=

- \frac{20.2}{( a _{0} - 8 ) ^{2}}

and passing through

(a_{0}, \frac{20.2}{a _{0} - 8}) : y = - \frac{20.2}{( a _{0} - 8 ) ^{2}} x + \frac{40.4 a _{0} - 161.6}{( a _{0} - 8 ) ^{2}}

y = - \frac{20.2}{( a _{0} - 8 ) ^{2}} x + \frac{40.4 a _{0} - 161.6}{( a _{0} - 8 ) ^{2}}

\int r 1 - r^{2} d r

7 x^{2} - 3 x, x^{3} - 7 x + 2

x \to πk lim (sin (\frac{1}{x}))

y^{^{'}} - \frac{2}{x} y = 2 x^{- 1} y^{- 1}

\int x e^{y + 2 z} d y

What is the tangent of 2/(x-8)(10.1) ?
The tangent of 2/(x-8)(10.1) is y=-(20.2)/((a_{0)-8)^2}x+(40.4a_{0}-161.6)/((a_{0)-8)^2}