# Differentiate the following function

$$y = \sqrt {\sin x} = (\sin x)^{\frac 12}$$

\begin{aligned} {dy \over dx} & = \frac 12 (\sin x)^{-\frac {1}2}{d\over dx} \sin x \\ & = \frac 12 (\sin x)^{-\frac 12} \cos x \\ & = \frac{\cos x}{ 2\sqrt{\sin x}} \end{aligned}

Is this correct?

Please excuse the poor layout, i'm new :'(

• Please use $\frac{dy}{dx}$ for the derivative... – user88595 Jun 1 '14 at 10:59
• See my answer below, and by the way, welcome aboard! :) – DanielY Jun 1 '14 at 11:01

Yes my friend, you are correct!

For future derivative checks, try www.wolframalpha.com :)

• +1 for mentioning Wolfram Alpha, very handy sometimes ^^ – Jori Jun 1 '14 at 18:44

$$\frac{d}{dx}\sqrt{\sin x} = \frac{1}{2 \sqrt{\sin x}} \frac{d}{dx}\sin x = \frac{\cos x}{2 \sqrt{\sin x}}$$ by the chain rule.

• I think that's exactly what the OP found, in his post. No need to redo work done correctly. – Namaste Jun 1 '14 at 11:42

1) If $y = \sin(x)^{1/2}$ then $y \neq \dfrac{1}{2}\sin(x)^{-1/2}\dfrac{d}{dx}\sin(x)$ (a function and its derivative are two different things)

2) If $y = \dfrac{1}{2}\sin(x)^{-1/2}\dfrac{d}{dx}\sin(x)$ then $y \neq \dfrac{1}{2}\sin(x)^{1/2}\cos(x)$ (minus sign is missing)

You result is correct but your redaction is wrong.