How do you find the derivative of 1/(x^2-1)1x21 using the limit definition?

1 Answer
Sep 19, 2016

(df)/(dx)=(-2x)/(x^2-1)^2dfdx=2x(x21)2

Explanation:

as f(x)=1/(x^2-1)f(x)=1x21

f(x+h)=1/((x+h)^2-1)f(x+h)=1(x+h)21

Hence, f(x+h)-f(x)=1/((x+h)^2-1)-1/(x^2-1)f(x+h)f(x)=1(x+h)211x21

= ((x^2-1)-((x+h)^2-1))/((x^2-1)((x+h)^2-1))(x21)((x+h)21)(x21)((x+h)21)

= ((x^2-1)-(x^2+2hx+h^2-1))/((x^2-1)((x+h)^2-1))(x21)(x2+2hx+h21)(x21)((x+h)21)

= ((x^2-1-x^2-2hx-h^2+1))/((x^2-1)((x+h)^2-1))(x21x22hxh2+1)(x21)((x+h)21)

= ((-2hx-h^2))/((x^2-1)((x+h)^2-1))(2hxh2)(x21)((x+h)21) and

(f(x+h)-f(x))/h=((-2x-h))/((x^2-1)((x+h)^2-1))f(x+h)f(x)h=(2xh)(x21)((x+h)21)

Now (df)/(dx)=Lt_(h->0)(f(x+h)-f(x))/hdfdx=Lth0f(x+h)f(x)h

= Lt_(h->0)((-2x-h))/((x^2-1)((x+h)^2-1))Lth0(2xh)(x21)((x+h)21)

= (-2x)/((x^2-1)(x^2-1))2x(x21)(x21)

= (-2x)/(x^2-1)^22x(x21)2