How do you find an equation of the tangent line to the curve at the given point: #y=(2x)/(x+1)^2# at point (0,0)?

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Answer 1 · 2015-05-15T13:51:33

Find an equation of the tangent line to the curve at the given point: #y=(2x)/(x+1)^2# at point (0,0)

Find #y'# using the quotient rule, the power rule, and the chain rule:

#y'=((2)(x+1)^2-(2x)(2(x+1)*(1)))/(x+1)^4#

# =(2(x+1)^2-4x(x+1))/(x+1)^4#

We could do some more algebra, but it is clear at this point that when #x=0#, we have

#y'=2/1=2#

So the tangent line is through #(0,0)# and has slope #2#:

#y=2x#