AP Calculus AB and BC

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

Q: What is the derivative of sin(sqrt x)?

It is \frac{\cos\sqrt{x}}{2\sqrt{x}} for x>0. The chain rule gives \cos\sqrt{x} times the derivative of \sqrt{x}, which is \frac{1}{2\sqrt{x}}.

Q: Is sin(sqrt x) differentiable at x = 0?

No. As x\to 0^{+} the numerator \cos\sqrt{x} tends to 1 while 2\sqrt{x} tends to 0, so the slope grows without bound and the graph meets the origin with a vertical tangent.

The derivative of sin(sqrt x) is cos(sqrt x) divided by 2 sqrt x. The outer function is sine and the inner function is sqrt x, so the decisive fact is that the inner derivative is 1 over 2 sqrt x, a factor that lands in the denominator instead of out front.

\frac{d}{dx}\left[\sin\sqrt{x}\right] = \frac{\cos\sqrt{x}}{2\sqrt{x}}

Chain rule on sin(sqrt x)

The outer function is $\sin u$ and the inner function is $u=\sqrt{x}$ . Differentiating the outer gives $\cos u$ , and the inner derivative is $u'=\frac{1}{2\sqrt{x}}$ .

\frac{d}{dx}\sin\sqrt{x} = \cos\sqrt{x}\cdot\frac{1}{2\sqrt{x}} = \frac{\cos\sqrt{x}}{2\sqrt{x}}

If the derivative of the root is not automatic yet, rewrite $\sqrt{x}$ as $x^{1/2}$ . The power rule gives $\frac{1}{2}x^{-1/2}$ , which is $\frac{1}{2\sqrt{x}}$ . The formula holds for $x>0$ , since the root has no derivative at $0$ and no real values to the left of it.

Why the chain factor divides here

Compare this with $\sin^{2}x$ , where the chain factor multiplies rather than divides. The rule being used is identical in both cases: multiply by $u'$ . What changes is what $u'$ happens to be, $\cos x$ there and $\frac{1}{2\sqrt{x}}$ here.

\frac{d}{dx}\sin^{2}x = 2\sin x\cdot\cos x, \qquad \frac{d}{dx}\sin\sqrt{x} = \cos\sqrt{x}\cdot\frac{1}{2\sqrt{x}}

Multiplying by $\frac{1}{2\sqrt{x}}$ is the same as dividing by $2\sqrt{x}$ , so no new rule is in play. The effect on the graph is real, though: that factor shrinks as $x$ grows, so the wiggles of $\sin\sqrt{x}$ flatten out and stretch, while the wiggles of $\sin^{2}x$ keep the same steepness forever.

The mistakes students make

Three wrong answers turn up far more often than any others on this derivative.

Answering $\cos\sqrt{x}$ and stopping. That is the outer derivative on its own, with the chain factor $\frac{1}{2\sqrt{x}}$ left out entirely.
Answering $2\sqrt{x}\cos\sqrt{x}$ , flipping the chain factor upside down and multiplying by $2\sqrt{x}$ instead of dividing by it. The root belongs underneath.
Answering $\frac{\cos x}{2\sqrt{x}}$ , evaluating the cosine at $x$ instead of at $\sqrt{x}$ . The argument never changes when you differentiate the outer function.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of sin(sqrt x)?

It is $\frac{\cos\sqrt{x}}{2\sqrt{x}}$ for $x>0$ . The chain rule gives $\cos\sqrt{x}$ times the derivative of $\sqrt{x}$ , which is $\frac{1}{2\sqrt{x}}$ .

Why is the 2 sqrt x on the bottom instead of in front?

Because the inner derivative itself is a fraction. The chain rule multiplies by $u'=\frac{1}{2\sqrt{x}}$ , and multiplying by that reciprocal is the same as dividing by $2\sqrt{x}$ .

Is sin(sqrt x) differentiable at x = 0?

No. As $x\to 0^{+}$ the numerator $\cos\sqrt{x}$ tends to $1$ while $2\sqrt{x}$ tends to $0$ , so the slope grows without bound and the graph meets the origin with a vertical tangent.