What is the tangent of f(x)= 6/(x-1) ?

The tangent of f(x)= 6/(x-1) is y=-6/((a_{0)-1)^2}x+(12a_{0}-6)/((a_{0)-1)^2}

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tangent of f(x)= 6/(x-1)

tangent of

f (x) = \frac{6}{x - 1}

y = - \frac{6}{( a _{0} - 1 ) ^{2}} x + \frac{12 a _{0} - 6}{( a _{0} - 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{6}{a _{0} - 1})

Find the slope of

f (x) = \frac{6}{x - 1} : \frac{df ( x )}{d x} = - \frac{6}{( 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{6}{( a _{0} - 1 ) ^{2}}

Find the line with slope m=

- \frac{6}{( a _{0} - 1 ) ^{2}}

and passing through

(a_{0}, \frac{6}{a _{0} - 1}) : y = - \frac{6}{( a _{0} - 1 ) ^{2}} x + \frac{12 a _{0} - 6}{( a _{0} - 1 ) ^{2}}

y = - \frac{6}{( a _{0} - 1 ) ^{2}} x + \frac{12 a _{0} - 6}{( a _{0} - 1 ) ^{2}}

\int (4 x^{3} + 1) d x

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

\int 3 x^{2} (2 x^{2} + \frac{1}{x ^{3}}) d x

\frac{d}{d x} (16 - x^{\frac{2}{3}})

cos ((π x)^{6})

What is the tangent of f(x)= 6/(x-1) ?
The tangent of f(x)= 6/(x-1) is y=-6/((a_{0)-1)^2}x+(12a_{0}-6)/((a_{0)-1)^2}