What is the tangent of (7x)/(x-1) ?

The tangent of (7x)/(x-1) is y=-7/((a_{0)-1)^2}x+(7a_{0}^2)/((a_{0)-1)^2}

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tangent of (7x)/(x-1)

tangent of

\frac{7 x}{x - 1}

y = - \frac{7}{( a _{0} - 1 ) ^{2}} x + \frac{7 a _{0}^{2}}{( a _{0} - 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{7 a _{0}}{a _{0} - 1})

Find the slope of

y = \frac{7 x}{x - 1} : \frac{d y}{d x} = - \frac{7}{( x - 1 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{7}{( a _{0} - 1 ) ^{2}}

Find the line with slope m=

- \frac{7}{( a _{0} - 1 ) ^{2}}

and passing through

(a_{0}, \frac{7 a _{0}}{a _{0} - 1}) : y = - \frac{7}{( a _{0} - 1 ) ^{2}} x + \frac{7 a _{0}^{2}}{( a _{0} - 1 ) ^{2}}

y = - \frac{7}{( a _{0} - 1 ) ^{2}} x + \frac{7 a _{0}^{2}}{( a _{0} - 1 ) ^{2}}

\int_{0}^{\infty} x e^{- x y} d x

(x) (3 - x)^{2}

\frac{d}{d x} (y cos (x ln (y)))

\frac{x}{2 - x}

\frac{d y}{d x} = 5 (1 - \frac{x}{10})

What is the tangent of (7x)/(x-1) ?
The tangent of (7x)/(x-1) is y=-7/((a_{0)-1)^2}x+(7a_{0}^2)/((a_{0)-1)^2}