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

The tangent of y= 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 y= 2/(x-8)(10.1)

tangent of

y = \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}}

n = 1 \sum \infty \frac{1}{n ^{\frac{1}{3}}}

\frac{3 ( 1 - 4 x )}{58 ( x ^{2} + 9 )}

sin (ln (cos (t)))

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