Question

Can anyone guide me on how to do this , :)

Answer 1

\[\huge y = \sqrt{\sin \sqrt{x}}\]

Answer 2

Differentiate this

Answer 3

Use chain rule

Answer 4

Rewriting with rational exponents might help:
\[\sqrt{\sin\sqrt x}=\left(\sin x^{1/2}\right)^{1/2}\]
Then chain rule.

Answer 5

It states that:

\[\large{\cfrac{df(g(x))}{dx} = f'(g(x)).g'(x).1}\]

Answer 6

\[y=\left(\sin x^{1/2}\right)^{1/2}\]
\[y'=\frac{1}{2}\left(\sin x^{1/2}\right)^{-1/2}\cdot\cos x^{1/2}\cdot \frac{1}{2}x^{-1/2}\]

Answer 7

Outside inside rule , yes

Answer 8

IS that the answer

Answer 9

Well, you can still simplify/rewrite at your leisure.

Answer 10

Yes

Answer 11

oh i see thanks

Answer 12

You can similarly try differentiating other functions.

Answer 13

You can derive the quotient rule of differentiation using chain rule

Answer 14

Want to give it a try ?

Answer 15

Yes sure :)
How to enter it into wolfram

Answer 16

Do you want to verify the answer ?

Answer 17

http://www.wolframalpha.com/input/?i=D%5BSqrt%5BSin%5BSqrt%5Bx%5D%5D%5D%2Cx%5D

Answer 18

yes

Answer 19

Well there you go .. @SithsAndGiggles has provided the link :)