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}}

Compute the slope of

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

x = a_{0} : 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}}

x \to + 1 + lim (\frac{7}{1 - x})

\int \frac{27}{x ^{2}} d x

\int \frac{2}{x ( x - 2 ) ^{2}} d x

\frac{d}{d x} (x^{3} cos (y))

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

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}