Find equation of tangetn line to the curve

Question

I need to see the steps for solving the following equation:  Find the equation of the tangent line to the curve at the given points for y=x^2-1/x^2+x+1  (1,0).  I know the answer is 2/3x-2/3 but I'm just not getting that when I try to derive the equation and then plug that answer into the line formula.

Answer 1

Find the equation of the tangent line to
f(x)=(x^2-1)/(x^2+x+1) at the point (1,0).

m = f’(1)

Use Quotient Rule:

f’(x) = ((x^2-1)’(x^2+x+1)–(x^2-1)(x^2+x+1)’)
/((x^2+x+1)^2)

f’(x) = (2x(x^2+x+1)–(x^2-1)(2x+1))
/((x^2+x+1)^2)

m = f’(1) = (2(3)–0)/(3^2) = 6/9 = 2/3

Tangent Line:

y–0 = 2/3 (x–1)

y = 2/3 x - 2/3

Answer 2

So to get the slope at (1,0) you need to find derivative of equation

 

y= x² - 1/x² + x + 1

 

Rewrite 1/x² as x^-2

 

so

y= x² - x^-2 + x + 1

 

do derivative

 

y' = 2x+2x^-3 + 1

 

To get slope plug 1 into the x

 

y'(1)= 2(1)+2(1^-3) + 1

 

y'(1)= 2+2+1

y'(1)=5

 

y=5(x-1)