Derivative of sin x/(1+cos x): Answer and Proof

Quotient rule, then the Pythagorean identity

Everything turns on one minus sign. The numerator $\sin x$ differentiates to $\cos x$, but the denominator $1+\cos x$ differentiates to $-\sin x$, and that negative meets the subtraction built into the quotient rule. Two minus signs make a plus.

Apply $\sin^{2}x+\cos^{2}x = 1$ to the last two terms. The numerator becomes $1+\cos x$, an exact copy of one factor in the denominator, so it cancels.

The same function written as tan(x/2)

There is a reason the derivative is tidier than the function: the function is $\tan\frac{x}{2}$ in disguise. Two standard identities, $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1+\cos x = 2\cos^{2}\frac{x}{2}$, reduce the quotient in one step.

Differentiating $\tan\frac{x}{2}$ by the chain rule gives $\frac{1}{2}\sec^{2}\frac{x}{2}$. Since $1+\cos x = 2\cos^{2}\frac{x}{2}$, that expression is $\frac{1}{2\cos^{2}\frac{x}{2}} = \frac{1}{1+\cos x}$. Two independent routes, one answer.

Wherever it is defined, $1+\cos x > 0$, so $f' > 0$ and the function increases on each interval between the points $x = \pi+2k\pi$ where the denominator vanishes. The sample inputs $x = -1, 0.5, 2$ all lie in $\left(-\pi,\pi\right)$.

The mistakes students make

The first error stops the identity from firing, the second throws away a factor after the numerator is already correct, and the third abandons the quotient rule altogether.

• Writing the numerator as $\cos x+\cos^{2}x-\sin^{2}x$. The quotient rule subtracts $\sin x\cdot\left(-\sin x\right)$, which is $+\sin^{2}x$. With a minus there the terms never combine to $1$, and the answer comes out as $\frac{\cos x+\cos 2x}{\left(1+\cos x\right)^{2}}$.
• Cancelling the whole denominator and answering $1$. Only one of the two factors of $1+\cos x$ is matched by the numerator, so one factor remains and the answer is $\frac{1}{1+\cos x}$.
• Dividing derivative by derivative and answering $\frac{\cos x}{-\sin x} = -\cot x$. That is not the quotient rule, and on $\left(0,\frac{\pi}{2}\right)$ it even comes out negative, while the true slope $\frac{1}{1+\cos x}$ is positive wherever the function is defined.