What is the tangent of (6x)/((x+1)) ?

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

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tangent of (6x)/((x+1))

tangent of

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

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

Find the slope of

y = \frac{6 x}{x + 1} : \frac{d y}{d x} = \frac{6}{( x + 1 ) ^{2}}

Compute the slope of

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

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

\int \frac{7 x ^{2} + x + 54}{x ^{3} + 9 x} d x

y = x + 2

\frac{d}{d x} (\frac{5}{25 x ^{2} + 1})

\int \frac{2 t ^{5}}{5} - \frac{2 t ^{2}}{5} - \frac{2 t}{5} d t

\frac{d}{d x} (ln (3^{3 x}))

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