nirmalnema
  • nirmalnema
what is the differentiation of sq rt of cos sq rt x.?
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\sqrt{\cos(\sqrt{x})}\] ?
nirmalnema
  • nirmalnema
yes..plz explain the whole process
anonymous
  • anonymous
you need the chain rule, three times

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anonymous
  • anonymous
this is a composition of functions it is the 1) square root of the 2) cosine of the 3) square root
nirmalnema
  • nirmalnema
yup i no that but i dont know how to use it
anonymous
  • anonymous
so before you start, you need to know two facts the derivative of the square root of something is one over two times the square root of something, and the derivative of the cosine of something is minus the sign of something
anonymous
  • anonymous
sine
anonymous
  • anonymous
more commonly written as \[\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}\] and \[\frac{d}{dx}[\cos(x)]=-\sin(x)\]
anonymous
  • anonymous
so first we take care of the outer most function, which is the square root function step one is \[\frac{d}{dx}[\sqrt{\cos(\sqrt{x})}]=\frac{1}{2\sqrt{\cos(\sqrt{x})}}\times \frac{d}{dx}[\cos(\sqrt{x})]\]
nirmalnema
  • nirmalnema
okk...
anonymous
  • anonymous
in other words, it is one over two times the square root of what was inside the radical the next step is to take the derivative of cosine, which is minus sine
anonymous
  • anonymous
\[\frac{d}{dx}[\sqrt{\cos(\sqrt{x})}]=\frac{1}{2\sqrt{\cos(\sqrt{x})}}\times \frac{d}{dx}[\cos(\sqrt{x})]\] \[=\frac{1}{2\sqrt{\cos(\sqrt{x})}}\times -\sin(\sqrt{x})\times \frac{d}{dx}\sqrt{x}\]
anonymous
  • anonymous
the last step is to take the derivative of \(\sqrt{x}\) and you end up with \[=\frac{1}{2\sqrt{\cos(\sqrt{x})}}\times -\sin(\sqrt{x})\times \frac{1}{2\sqrt{x}}\]
anonymous
  • anonymous
you can clean this up a bit and write \[-\frac{\sin(\sqrt{x})}{4\sqrt{x}\cos(\sqrt{x})}\]
anonymous
  • anonymous
it is really not as bad as it looks, i just wrote down a lot of steps to show what was going on
nirmalnema
  • nirmalnema
ok thanks..
anonymous
  • anonymous
yw

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