What is the tangent of ((x+5))/((x-5)) ?

The tangent of ((x+5))/((x-5)) is y=-(10)/((a_{0)-5)^2}x+(a_{0}+5)/(a_{0)-5}+(10a_{0})/((a_{0)-5)^2}

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tangent of ((x+5))/((x-5))

tangent of

\frac{( x + 5 )}{( x - 5 )}

y = - \frac{10}{( a _{0} - 5 ) ^{2}} x + \frac{a _{0} + 5}{a _{0} - 5} + \frac{10 a _{0}}{( a _{0} - 5 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0} + 5}{a _{0} - 5})

Find the slope of

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

Compute the slope of

y = \frac{x + 5}{x - 5}

x = a_{0} : m = - \frac{10}{( a _{0} - 5 ) ^{2}}

Find the line with slope m=

- \frac{10}{( a _{0} - 5 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0} + 5}{a _{0} - 5}) : y = - \frac{10}{( a _{0} - 5 ) ^{2}} x + \frac{a _{0} + 5}{a _{0} - 5} + \frac{10 a _{0}}{( a _{0} - 5 ) ^{2}}

y = - \frac{10}{( a _{0} - 5 ) ^{2}} x + \frac{a _{0} + 5}{a _{0} - 5} + \frac{10 a _{0}}{( a _{0} - 5 ) ^{2}}

f (x) = x^{2} - tan (x), at x = 0

\frac{\partial}{\partial y} (9 e^{yz} cos (x yz))

\int \frac{3 ln ^{2} ( x ) + 25}{x} d x

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

sin (x cos (x))

What is the tangent of ((x+5))/((x-5)) ?
The tangent of ((x+5))/((x-5)) is y=-(10)/((a_{0)-5)^2}x+(a_{0}+5)/(a_{0)-5}+(10a_{0})/((a_{0)-5)^2}