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

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Answer 1 · 2016-09-19T06:49:29

$(df)/(dx)=(-2x)/(x^2-1)^2$

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

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

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

= $((x^2-1)-((x+h)^2-1))/((x^2-1)((x+h)^2-1))$

= $((x^2-1)-(x^2+2hx+h^2-1))/((x^2-1)((x+h)^2-1))$

= $((x^2-1-x^2-2hx-h^2+1))/((x^2-1)((x+h)^2-1))$

= $((-2hx-h^2))/((x^2-1)((x+h)^2-1))$ and

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

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

= $Lt_(h->0)((-2x-h))/((x^2-1)((x+h)^2-1))$

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

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

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