



 
First of all $x^2  1 = (x  1)(x + 1).$ When $x \ne 1$ the denominator of $f(x)$ is not zero so $f(x)$ simplifies to $x + 1$ which is differentiable. When $x = 1$ the denominator of $f(x)$ is zero so $f(1)$ is not defined. Thus $f(x)$ is not differentiable at $x = 1.$ Penny  


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