Differentiation Help: Find dy/dx of y=4-1/x

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To find dy/dx for the equation y=4-1/x, one must differentiate the function correctly. The initial confusion stemmed from misinterpreting the differentiation process, as the user mistakenly rewrote the equation instead of applying differentiation rules. The correct differentiation leads to dy/dx being expressed as x^-2, not x-2. Understanding that dy/dx represents the derivative of y with respect to x is crucial for solving such problems. Clarifying these concepts can help in accurately performing differentiation in the future.
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1. Hello everyone, my problem is with differentation. It seems to be quite simple but i can't understand it. Its basically find dy/dx of this equation:
y=4-1/x




2. The only relevant equation is dy/dx



3. I know how to differentate the equation to get y'=4-x-1 but i don't know how to get from there to dy/dx. To be honest I am not really sure what dy/dx is. I have been told the answer is dy/dx=x-2. Can someone please give me a clear explanation to what dy/dx means and explain to me how i get from the y' equation to dy/dx. Thanks for any replies :)
 
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first y'(x) is dy/dx, second y = 4 - x-1 to get to y' ie dy/dx you have to differentiate it but the only thing you did as I seen is rewrote y from 4-1/x to 4- x-1 that is not differentiation.
 
Don't worry i see where i was going wrong i understand now for some reason i thought Differentiate ment somthing else, thanks for replying anywayz
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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