How do you find the equation of the tangent line to the curve #f(x)=x^2/(x+1)# at (2, 4/3)?

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Answer 1 · 2016-03-16T16:16:02

Equation of tangent at #(2,4/3)# is #8x-9y=4#

#d/dx((f(x))/(g(x)))=(d/dx(f(x))xxg(x)-f(x)xxd/dx(g(x)))/((g(x))^2#

Hence, differential of #x^2/(x+1)# is

#(d/dx(x^2)xx(x+1)-x^2xxd/dx(x+1))/(x+1)^2# or

#(2x(x+1)-x^2xx1)/(x+1)^2# or #(2x(x+1)-x^2)/(x+1)^2# or

#d/dx(x^2/(x+1))=(x^2+2x)/(x+1)^2=(x(x+2))/(x+1)^2#

As differential gives the slope of tangent at a point, slope of tangent at #(2,4/3)# is #(2xx(2+2))/(2+1)^2# or #8/9#.

Hence equation of tangent at #(2,4/3)# is

#(y-4/3)=8/9(x-2)# or #9(y-4/3)=8(x-2)# or

#9y-12=8x-16# or

#8x-9y=4#