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?
 A: You need the chain rule:
$$
\frac d {dx} \cos(x^2-1) = -\sin (x^2-1)\cdot\frac d{dx}(x^2-1)=\cdots.
$$
A: 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.
A: Differentiating $\cos(x^2-1)$ gives you $-2x\sin(x^2-1)$, using the Chain Rule
A: $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}\Longrightarrow$$
$$\frac{df(x)}{dx}\left(\frac{\cos{(x^2 - 1)}}{2x}\right)=$$
$$\frac{1}{2}\left(\frac{df(x)}{dx}\left(\frac{\cos{(x^2 - 1)}}{x}\right)\right)=$$
$$\frac{1}{2}\left(\frac{x\frac{df(x)}{x}(\cos(1-x^2))-\cos(1-x^2)\frac{df(x)}{dx}x}{x^2}\right)=$$
$$\frac{x\frac{df(x)}{x}(\cos(1-x^2))-1\cos(1-x^2)}{2x^2}=$$
$$\frac{-\cos(1-x^2)+2x(x\sin(1-x^2))}{2x^2}=$$
$$\frac{-\cos(1-x^2)+x^2\sin(1-x^2)}{2x^2}$$
