# Search results

1. ### Different results with quotient rule

You see the minus sign in front of \left( x^2 -1 \right) \cdot 1, right? That is just short notation for - \left( x^2 -1 \right) \cdot 1=\left( -1 \right) \cdot \left( x^2 -1 \right)= (-1) \cdot x^2 + (-1) \cdot (-1) = -x^2+1 Agreed? Or are you pulling my leg here? EDIT: Ah, good. I was...
2. ### Different results with quotient rule

It was also multiplied by -1, from the quotient rule formula.
3. ### Different results with quotient rule

Alright, for a function f(x) = \frac{g(x)}{h(x)} the quotient rule says f'(x) = \frac{g'(x)h(x)-g(x)h'(x)}{h^2(x)} In your case, g(x)=x^2-1 and h(x)=x. Thus g'(x)=2x and h'(x)=1. Then you get \frac{\mathrm{d}}{\mathrm{dx}}\frac{x^2-1}{x}=\frac{2x\cdot x - \left( x^2 -1 \right) \cdot...
4. ### Different results with quotient rule

I get the same answer using both methods. Are you sure you include all terms in the quotient rule? http://en.wikipedia.org/wiki/Quotient_rule
5. ### Why is this function injective?

That is right.
6. ### Why is this function injective?

Have you tried comparing f(f(x)) and f(f(y))?