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What is the dy/dx of 1/x ?
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use the standard formula for x^n 1/x = x^(-1)
\[1/x= x^{-1}\] \[dy/dx x^{-1}=-x^{-2}\] Bring down the exponent and then subtract one from it :)
ofcourse its correct, you got the result -1/x^2, correct. just the method used is different
Don't forget that a negative (-) exponent can be removed, and the term flipped. They said \[-1/x^{2}\]which equals my answer of \[-x^{−2}\] Raising something to a negative exponent basically just flips it :) And @hartnn is right as well, they did it with the limit definition of a derivative, rather than shortcuts
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I see!! Thanks a lot !!
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