What is the tangent of ((5x-4))/(x-1) ?

The tangent of ((5x-4))/(x-1) is y=-1/((a_{0)-1)^2}x+(5a_{0}^2-8a_{0}+4)/((a_{0)-1)^2}

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tangent of ((5x-4))/(x-1)

tangent of

\frac{( 5 x - 4 )}{x - 1}

y = - \frac{1}{( a _{0} - 1 ) ^{2}} x + \frac{5 a _{0}^{2} - 8 a _{0} + 4}{( a _{0} - 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{5 a _{0} - 4}{a _{0} - 1})

Find the slope of

y = \frac{5 x - 4}{x - 1} : \frac{d y}{d x} = - \frac{1}{( x - 1 ) ^{2}}

Compute the slope of

y = \frac{5 x - 4}{x - 1}

x = a_{0} : m = - \frac{1}{( a _{0} - 1 ) ^{2}}

Find the line with slope m=

- \frac{1}{( a _{0} - 1 ) ^{2}}

and passing through

(a_{0}, \frac{5 a _{0} - 4}{a _{0} - 1}) : y = - \frac{1}{( a _{0} - 1 ) ^{2}} x + \frac{5 a _{0}^{2} - 8 a _{0} + 4}{( a _{0} - 1 ) ^{2}}

y = - \frac{1}{( a _{0} - 1 ) ^{2}} x + \frac{5 a _{0}^{2} - 8 a _{0} + 4}{( a _{0} - 1 ) ^{2}}

\frac{d}{d x} (- s t)

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

3 ln (x)

\int \frac{1}{x ( x + 1 ) ^{2}} d x

\frac{d}{d x} (e^{- 2 x} \cdot cos (3 x))

What is the tangent of ((5x-4))/(x-1) ?
The tangent of ((5x-4))/(x-1) is y=-1/((a_{0)-1)^2}x+(5a_{0}^2-8a_{0}+4)/((a_{0)-1)^2}