# consider the equation: f(x) = (x - 1) / (X^2 - 1) find f'(x) 5 marks for questionthe equation is (x - 1) divided by (x^2 - 1).

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

Let us try and simplify the denominator.

We know that : a^2 - b^2 = (a-b)(a+b)

In our case:

(x^2 - 1) = (x-1)(x+1)\

Now let us rewrite:

f(x) = (x-1)/(x-1)(x+1)

Now reduce similar:

==> f(x) = 1/(x+1)

Now by definition :

**f'(x) = -1/(x+1)^2 **

f(x) = (x - 1) / (X^2 - 1)

f(x)=(x-1)/(x-1)(x+1)

f(x)=1/(x+1)

f(x)=(x+1)^-1

f'(x)=[(-1)(x+1)^(-1-1)]*1

f'(x)=(-1)(x+1)^-2

f'(x)={-1/(x+1)^2}