What is the tangent of (6x)/((x+1)^2) ?

The tangent of (6x)/((x+1)^2) is y=(6(1-a_{0}))/((a_{0)+1)^3}x+(12a_{0}^2)/((a_{0)+1)^3}

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tangent of (6x)/((x+1)^2)

tangent of

\frac{6 x}{( x + 1 ) ^{2}}

y = \frac{6 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}} x + \frac{12 a _{0}^{2}}{( a _{0} + 1 ) ^{3}}

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 ) ^{2}})

Find the slope of

y = \frac{6 x}{( x + 1 ) ^{2}} : \frac{d y}{d x} = \frac{6 ( 1 - x )}{( x + 1 ) ^{3}}

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 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}}

Find the line with slope m=

\frac{6 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}}

and passing through

(a_{0}, \frac{6 a _{0}}{( a _{0} + 1 ) ^{2}}) : y = \frac{6 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}} x + \frac{12 a _{0}^{2}}{( a _{0} + 1 ) ^{3}}

y = \frac{6 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}} x + \frac{12 a _{0}^{2}}{( a _{0} + 1 ) ^{3}}

\frac{d}{d x} (5 x^{2} + 4 x^{7})

\frac{d}{d x} (\frac{2 ^{x}}{x} + x - lo g_{2} (x))

(- 19, - 6), (15, 16)

\int sin^{2} (4 x) cos (4 x) d x

\frac{\partial}{\partial y} (y cos (x))

What is the tangent of (6x)/((x+1)^2) ?
The tangent of (6x)/((x+1)^2) is y=(6(1-a_{0}))/((a_{0)+1)^3}x+(12a_{0}^2)/((a_{0)+1)^3}