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find the derivatives d/dx(sqrt(xcosx))
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\[d/dx \sqrt{xcox}\]
You need product rule followed by chain rule
yes i used that but i made mistake but i dont know where!
What do you have?
\[f(x)=(xcosx)^{1/2}\] \[f(x)=1/2(xcosx)^{-1/2}\] \[f(x)=1/2(xcosx)^{1/2} d/dx (xcosx)\] ?
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\[f(x)=1/2(xcosx)^{1/2} xsinx?\]
The third line is \[f(x)=1/2(xcosx)^{-1/2} d/dx (xcosx)\]
And as for d/dx(xcosx) you have to use product rule \[y'=\frac{1}{2}(xcos(x))^{-1/2}[x(-\sin(x)+\cos(x)]\]
i dont understand xcosx part!
I gotta go now
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Product rule
Product rule is \[y=uv \\ \\ y'=udv+vdu\]
I gotta go bye
\[\frac{d}{dx}\sqrt{f(x)}=\frac{f'(x)}{2\sqrt{f(x)}}\] with \[f(x)=x\cos(x), f'(x)=\cos(x)-x\sin(x)\]
@satellite73 theres not root?
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