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OpenStudy (anonymous):

Find the derivative of

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

chainz :>

OpenStudy (anonymous):

yea i know that is chain rule

OpenStudy (anonymous):

obviously but i have to apply it here @tkhunny

OpenStudy (tkhunny):

Sure you do. What's this \(\dfrac{d}{dx}\sqrt{x} =\;\)??

OpenStudy (anonymous):

Okay...somehow, you have to pinpoint what I like to call the outer shell function :) The function... that covers all the others...

OpenStudy (anonymous):

I think THIS part of the function looks like a shell... don't you? :) \[\Large \color{red}{\sqrt{\color{black}{\sin \sqrt x}}}\]

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Well, here's how it works... just differentiate the 'shell' as if whatever's inside is just a big fancy x...first.... now, what is the derivative of 'square root'? ^^

OpenStudy (anonymous):

of root x ?

OpenStudy (anonymous):

yeah, that...

OpenStudy (anonymous):

1/2root x

OpenStudy (tkhunny):

Okay, so the first piece is \(\dfrac{1}{2\sqrt{sin(\sqrt{x})}}\). What's next?

OpenStudy (anonymous):

^okay, as much as I'd like to say that's it, that's not it :D After you differentiate the 'outer shell', you then multiply that entire expression by the derivative of WHATEVER WAS INSIDE that shell thingy... So inside the outermost radical is a \(\large \sin \sqrt x\) So what's the derivative of that? (lawlz chain rule again)

OpenStudy (anonymous):

|dw:1404497002056:dw|

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