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

Question

William Abdalla · Accepted Answer

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

According to quotient rule

#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#

Elijah Abram · Answer

undefined To find the equation of the tangent line to the curve f(x) = x^2/(x+1) at the point (2, 4/3), we need to find the slope of the tangent line and then use the point-slope form of a linear equation.

1. Find the derivative of f(x) using the quotient rule:
   f'(x) = [(x+1)(2x) - x^2(1)] / (x+1)^2

2. Simplify the derivative:
   f'(x) = (2x^2 + 2x - x^2) / (x+1)^2
          = (x^2 + 2x) / (x+1)^2

3. Evaluate the derivative at x = 2 to find the slope of the tangent line:
   f'(2) = (2^2 + 2(2)) / (2+1)^2
         = (4 + 4) / 9
         = 8/9

4. Use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope:
   y - (4/3) = (8/9)(x - 2)

5. Simplify the equation:
   y - 4/3 = 8/9(x - 2)

Thus, the equation of the tangent line to the curve f(x) = x^2/(x+1) at the point (2, 4/3) is y - 4/3 = 8/9(x - 2).