Derivative of x/sin x: Quotient Rule Proof, Mistakes

Applying the quotient rule

Write $u = x$ on top and $v = \sin x$ on the bottom, so $u' = 1$ and $v' = \cos x$. The quotient rule is $\frac{u'v-uv'}{v^{2}}$, and the subtraction runs in that order and no other.

Nothing cancels. The numerator $\sin x - x\cos x$ has no common factor with $\sin^{2}x$, so this is already the finished form. An equivalent version, obtained by splitting the fraction, is $\csc x - x\cot x\csc x$.

What happens at x = 0

$\frac{x}{\sin x}$ is the reciprocal of the famous limit $\frac{\sin x}{x}$, so the function is undefined at $x = 0$ but approaches $1$ as $x$ approaches $0$. There is a removable discontinuity there, not a vertical asymptote.

The derivative behaves the same way. Both $\sin x - x\cos x$ and $\sin^{2}x$ go to $0$ at the origin, and the ratio approaches $0$, which fits a graph that flattens out as it fills in the hole at height $1$. Real vertical asymptotes appear instead at $x = \pi, 2\pi$ and so on, where $\sin x = 0$ but $x$ does not.

The mistakes students make

The quotient rule punishes carelessness about order more than any other differentiation rule, and this function is a clean example of it.

• Writing $\frac{x\cos x-\sin x}{\sin^{2}x}$, the exact negative of the correct derivative. The rule starts with the bottom times the derivative of the top, so $\sin x$ comes first.
• Offering $\frac{1}{\cos x}$ by differentiating the top and bottom separately. That is not a differentiation rule. Differentiating top and bottom is L'Hopital's rule, and it returns the value of a limit at a single point, not a derivative function. It is doubly confusing here because at $x = 0$ the form really is $\frac{0}{0}$, so L'Hopital does apply there and gives $\frac{1}{\cos 0} = 1$, the limit found above, which is not the slope anywhere.
• Turning the denominator into $\sin x^{2}$ or $\sin\left(x^{2}\right)$. The squaring applies to the whole function $\sin x$, so it is $\left(\sin x\right)^{2}$, written $\sin^{2}x$.