Find the equation of the tangent line to the curve . . . ?

The general equation for slope is given by the first derivative. I would rewrite the original using a negative exponent and use the product rule to differentiate:

y = 2x(x+1)‾²

y' = m = 2(x+1)‾² + (2x)(-2)(x+1)‾³(1)

= 2(x+1)‾² - 4x(x+1)‾³

Solve this equation where x=0 to find the slope at that point:

m = 2(0+1)‾² - 4(0)(0+1)‾³

= 2(1)‾² - 0

= 2

Since the tangent line goes through the origin (0,0) you know its y-intercept (b) is 0 and you know its slope (m) from the equation you solved:

y = mx+b

= 2x + 0

y = 2x <---- that's your tangent line

y = 2x /(x + 1)^2

find y ' which is slope of tangent

y ' = [ (x+1)^2( 2) - 2x (x+1) ] / (x + 1)^4

y ' = 2(x + 1)[ x + 1 - x ] / (x + 1)^2

y ' = -2 / (x + 1)

at x = 0, y ' = -2 /1 = -2

The eqn of tangent with slope -2 and passing through (0, 0) is

y - 0 = -2 (x - 0)

y = -2x

y + 2x = 0 --------------eqn of tangent

f ` ( x ) , by quotient rule , is given by :-

( x + 1 ) ² (2) - (2x) (2) ( x + 1 )

---------------------------------------

( x + 1 )^4

( x + 1 ) ² (2) - (4x) ( x + 1 )

---------------------------------------

( x + 1 )^4

2 ( x + 1 ) [ ( x + 1 ) - 2x ]

---------------------------------------

( x + 1 )^4

2 ( x + 1 ) [ 1 - x ]

------------------------

( x + 1 )^4

2 ( 1 - x )

---------------

( x + 1 )^3

f ` ( 0 ) = 2 = m

y = 2x is equation of tangent line.