What is the tangent of (8x)/((x+1)^2) ?

The tangent of (8x)/((x+1)^2) is y=(8(1-a_{0}))/((a_{0)+1)^3}x+(16a_{0}^2)/((a_{0)+1)^3}

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tangent of (8x)/((x+1)^2)

tangent of

\frac{8 x}{( x + 1 ) ^{2}}

y = \frac{8 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}} x + \frac{16 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{8 a _{0}}{( a _{0} + 1 ) ^{2}})

Find the slope of

y = \frac{8 x}{( x + 1 ) ^{2}} : \frac{d y}{d x} = \frac{8 ( 1 - x )}{( x + 1 ) ^{3}}

Compute the slope of

y = \frac{8 x}{( x + 1 ) ^{2}}

x = a_{0} : m = \frac{8 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}}

Find the line with slope m=

\frac{8 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}}

and passing through

(a_{0}, \frac{8 a _{0}}{( a _{0} + 1 ) ^{2}}) : y = \frac{8 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}} x + \frac{16 a _{0}^{2}}{( a _{0} + 1 ) ^{3}}

y = \frac{8 ( 1 - a _{0} )}{( a _{0} + 1 ) ^{3}} x + \frac{16 a _{0}^{2}}{( a _{0} + 1 ) ^{3}}

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

\frac{d}{d x} (cos (\frac{1 - e ^{2 x}}{1 + e ^{2 x}}))

F (r) = \frac{7}{r ^{3}}

\int_{0}^{1} 5 ln (4 x) d x

x + 3, \frac{1}{2} x + 3

What is the tangent of (8x)/((x+1)^2) ?
The tangent of (8x)/((x+1)^2) is y=(8(1-a_{0}))/((a_{0)+1)^3}x+(16a_{0}^2)/((a_{0)+1)^3}