3
$\begingroup$

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

\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 :'(

$\endgroup$
  • 4
    $\begingroup$ Please use $\frac{dy}{dx}$ for the derivative... $\endgroup$ – user88595 Jun 1 '14 at 10:59
  • $\begingroup$ See my answer below, and by the way, welcome aboard! :) $\endgroup$ – DanielY Jun 1 '14 at 11:01
9
$\begingroup$

Yes my friend, you are correct!

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

$\endgroup$
  • $\begingroup$ +1 for mentioning Wolfram Alpha, very handy sometimes ^^ $\endgroup$ – Jori Jun 1 '14 at 18:44
2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ I think that's exactly what the OP found, in his post. No need to redo work done correctly. $\endgroup$ – Namaste Jun 1 '14 at 11:42
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.