Find y' for y = (sin x)/x + x/(sin x)

Find $y'$ for $y = \dfrac{\sin x}{x} + \dfrac{x}{\sin x}$.

Step 1: Differentiate the first term

By the quotient rule with $u = \sin x$, $v = x$: $\frac{d}{dx}\!\left(\frac{\sin x}{x}\right) = \frac{x\cos x - \sin x}{x^{2}}$.

Step 2: Differentiate the second term

By the quotient rule with $u = x$, $v = \sin x$: $\frac{d}{dx}\!\left(\frac{x}{\sin x}\right) = \frac{\sin x - x\cos x}{\sin^{2}x}$.

The two numerators are negatives of one another — a consequence of the terms being reciprocals.

Step 3: Combine and factor

Write the second numerator as $-\bigl(x\cos x-\sin x\bigr)$ and factor that common quantity out: \[y' = \frac{x\cos x-\sin x}{x^{2}} - \frac{x\cos x - \sin x}{\sin^{2}x} = \bigl(x\cos x-\sin x\bigr)\!\left(\frac{1}{x^{2}} - \frac{1}{\sin^{2}x}\right).\]

Step 4: Verification

\[x = \frac{\pi}{2}:\quad x\cos x - \sin x = 0 - 1 = -1,\quad \frac{1}{x^{2}} - \frac{1}{\sin^{2}x} = \frac{4}{\pi^{2}} - 1 \approx -0.59472\]

\[y' \approx (-1)(-0.59472) = 0.59472; \qquad \frac{y(\pi/2+10^{-5})-y(\pi/2-10^{-5})}{2\times 10^{-5}} \approx 0.59472 \;\;\checkmark\]

Step 5: Behaviour near the origin

At $x = 0$ both terms are undefined, but writing $s = \dfrac{\sin x}{x}$ gives $y = s + \dfrac{1}{s}$, and $s \to 1$ as $x \to 0$, so $y \to 2$: a removable singularity. The factored derivative confirms it — both factors vanish to third order and second order respectively, and their product tends to $0$.