1 Answer | Add Yours
First we will find the derivative of y.
==> We will use the quotient rule.
`==gt y' = ((x^2-1)'(x^2+x+1)- (x^2-1)(x^2+x+1)')/(x^2+x+1)^2 `
`==gt y' = (2x(x^2+x+1)- (2x+1)(x^2-1))/(x^2+x+1)^2 `
`==gt y'= (2x^3 +2x^2 + 2x - 2x^3 +2x +x^2 -1)/(x^2+x+1)^2 `
`==gt y' = (3x^2 + 4x -1)/(x^2+x+1)^2`
==> Now we subsitute with x = 1
`==gt y'(1)= (3+4-1)/(1+1+1)^2 = 6/9 = 2/3`
Then, the slope of the tangent line is m= `2/3`
Now we will find the equation of the tangent line where the slope is m= 2/3 and the point (1,0).
`==gt y-y1= m(x-x1) `
`==gt y= 2/3(x-1)`
`==gt y= 2/3 x - 2/3`
We’ve answered 317,664 questions. We can answer yours, too.Ask a question