Derivative of sqrt(sin x): Answer, Proof, Mistakes

The proof: chain rule on (sin x) to the one half

Rewrite the root as a power before differentiating: $\sqrt{\sin x} = (\sin x)^{\frac{1}{2}}$. The outer function is $u^{\frac{1}{2}}$ and the inner function is $u = \sin x$.

The power rule drops the exponent to $-\frac{1}{2}$, which is what moves the root into the denominator. The trailing $\cos x$ is the inner derivative, and leaving it out is the single most common error on this problem.

Domain and the vertical tangents

The function needs $\sin x \geq 0$, so it exists on $[0, \pi]$ and on every $2\pi$ shift of that interval. The derivative needs the strict inequality $\sin x > 0$, since $\sqrt{\sin x}$ is underneath.

At $x = 0$ the numerator approaches 1 while the denominator approaches 0, so the slope runs to $+\infty$ and the curve leaves the axis with a vertical tangent. The mirror image happens at $x = \pi$, where the slope runs to $-\infty$.

In between, the derivative is zero once, at $x = \frac{\pi}{2}$, where $\cos x = 0$. That is the maximum, and the value there is $\sqrt{1} = 1$.

Common mistakes

• Answering $\frac{1}{2\sqrt{\sin x}}$, which applies the power rule to the root but never multiplies by the inner derivative $\cos x$.
• Answering $\frac{\cos x}{2\sqrt{\cos x}}$, differentiating the inside and then putting the result back under the root. The root keeps the original inner function $\sin x$.
• Writing $\frac{\cos x}{2}\sqrt{\sin x}$. The exponent becomes negative, so the root belongs in the denominator, not the numerator.
• Confusing $\sqrt{\sin x}$ with $\sin\sqrt{x}$, whose derivative is $\frac{\cos\sqrt{x}}{2\sqrt{x}}$.