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Mathematics 18 Online
OpenStudy (anonymous):

find the derivatives d/dx(sqrt(xcosx))

OpenStudy (anonymous):

\[d/dx \sqrt{xcox}\]

sam (.sam.):

You need product rule followed by chain rule

OpenStudy (anonymous):

yes i used that but i made mistake but i dont know where!

sam (.sam.):

What do you have?

OpenStudy (anonymous):

\[f(x)=(xcosx)^{1/2}\] \[f(x)=1/2(xcosx)^{-1/2}\] \[f(x)=1/2(xcosx)^{1/2} d/dx (xcosx)\] ?

OpenStudy (anonymous):

\[f(x)=1/2(xcosx)^{1/2} xsinx?\]

sam (.sam.):

The third line is \[f(x)=1/2(xcosx)^{-1/2} d/dx (xcosx)\]

sam (.sam.):

And as for d/dx(xcosx) you have to use product rule \[y'=\frac{1}{2}(xcos(x))^{-1/2}[x(-\sin(x)+\cos(x)]\]

OpenStudy (anonymous):

i dont understand xcosx part!

sam (.sam.):

I gotta go now

sam (.sam.):

Product rule

sam (.sam.):

Product rule is \[y=uv \\ \\ y'=udv+vdu\]

sam (.sam.):

I gotta go bye

OpenStudy (anonymous):

\[\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\] with \[f(x)=x\cos(x), f'(x)=\cos(x)-x\sin(x)\]

OpenStudy (anonymous):

@satellite73 theres not root?

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