What is the tangent of ((x+3))/((x-3)) ?

The tangent of ((x+3))/((x-3)) is y=-6/((a_{0)-3)^2}x+(a_{0}+3)/(a_{0)-3}+(6a_{0})/((a_{0)-3)^2}

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tangent of ((x+3))/((x-3))

tangent of

\frac{( x + 3 )}{( x - 3 )}

y = - \frac{6}{( a _{0} - 3 ) ^{2}} x + \frac{a _{0} + 3}{a _{0} - 3} + \frac{6 a _{0}}{( a _{0} - 3 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0} + 3}{a _{0} - 3})

Find the slope of

y = \frac{x + 3}{x - 3} : \frac{d y}{d x} = - \frac{6}{( x - 3 ) ^{2}}

Compute the slope of

y = \frac{x + 3}{x - 3}

x = a_{0} : m = - \frac{6}{( a _{0} - 3 ) ^{2}}

Find the line with slope m=

- \frac{6}{( a _{0} - 3 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0} + 3}{a _{0} - 3}) : y = - \frac{6}{( a _{0} - 3 ) ^{2}} x + \frac{a _{0} + 3}{a _{0} - 3} + \frac{6 a _{0}}{( a _{0} - 3 ) ^{2}}

y = - \frac{6}{( a _{0} - 3 ) ^{2}} x + \frac{a _{0} + 3}{a _{0} - 3} + \frac{6 a _{0}}{( a _{0} - 3 ) ^{2}}

9 y^{^{''}} - y = x e^{\frac{x}{3}}, y (0) = 1, y^{^{'}} (0) = 0

\frac{\partial}{\partial y} (\frac{y}{x} + 4 x)

y = e^{x}, y = e^{- x}, x = - 2, x = 1

What is the tangent of ((x+3))/((x-3)) ?
The tangent of ((x+3))/((x-3)) is y=-6/((a_{0)-3)^2}x+(a_{0}+3)/(a_{0)-3}+(6a_{0})/((a_{0)-3)^2}