Question

what is the differentiation of sq rt of cos sq rt x.?

Answer 1

\[\sqrt{\cos(\sqrt{x})}\] ?

Answer 2

yes..plz explain the whole process

Answer 3

you need the chain rule, three times

Answer 4

this is a composition of functions
it is the
1) square root of the
2) cosine of the
3) square root

Answer 5

yup i no that but i dont know how to use it

Answer 6

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

Answer 7

sine

Answer 8

more commonly written as
\[\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}\] and
\[\frac{d}{dx}[\cos(x)]=-\sin(x)\]

Answer 9

so first we take care of the outer most function, which is the square root function
step one is
\[\frac{d}{dx}[\sqrt{\cos(\sqrt{x})}]=\frac{1}{2\sqrt{\cos(\sqrt{x})}}\times \frac{d}{dx}[\cos(\sqrt{x})]\]

Answer 10

okk...

Answer 11

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

Answer 12

\[\frac{d}{dx}[\sqrt{\cos(\sqrt{x})}]=\frac{1}{2\sqrt{\cos(\sqrt{x})}}\times \frac{d}{dx}[\cos(\sqrt{x})]\]
\[=\frac{1}{2\sqrt{\cos(\sqrt{x})}}\times -\sin(\sqrt{x})\times \frac{d}{dx}\sqrt{x}\]

Answer 13

the last step is to take the derivative of \(\sqrt{x}\) and you end up with \[=\frac{1}{2\sqrt{\cos(\sqrt{x})}}\times -\sin(\sqrt{x})\times \frac{1}{2\sqrt{x}}\]

Answer 14

you can clean this up a bit and write
\[-\frac{\sin(\sqrt{x})}{4\sqrt{x}\cos(\sqrt{x})}\]

Answer 15

it is really not as bad as it looks, i just wrote down a lot of steps to show what was going on

Answer 16

ok thanks..

Answer 17

yw