Derivative of $f(x) = \frac{\cos{(x^2 - 1)}}{2x}$ Find the derivative of the function $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}$$
This is my step-by-step solution: $$f'(x) = \frac{-\sin{(x^2 - 1)}2x - 2\cos{(x^2 -1)}}{4x^2} = \frac{2x\sin{(1 - x^2)} - 2\cos{(1 - x^2)}}{4x^2} = \frac{x\sin{(1 - x^2)} - \cos{(1 - x^2)}}{2x^2} = \frac{\sin{(1 - x^2)}}{2x} - \frac{\cos{(1-x^2)}}{2x^2}$$
and this is the output of WolphramAlpha: http://www.wolframalpha.com/input/?i=derivative+cos(x^2+-+1)%2F(2x)
Where is the mistakes?

In the first step, when you take the derivative of $\cos(x^2 - 1)$ in the quotient rule calculation, you're getting $-\sin(x^2-1)$, when you should be getting $-2x\sin(x^2 - 1)$ by the chain rule. This is then multiplied by the $2x$ from the denominator, giving $-4x^2\sin(x^2 - 1)$.
This is the only error--you'll notice that the difference between your solution and WolframAlpha's is the factor of $2x$ on the $\sin$ term.
