Question

Considering x^(a/n)= n-rt[x^(a)], wouldn't the 1/2 root of a number just be th square and the 1/3 the cube?

Answer 1

if you mean square root and cube root, then yes

Answer 2

No I dont mean \[\Large x^{\frac{1}{2}}\]=\[\Large \sqrt{x}\]


I mean \[\Huge \sqrt[1/2]{x}=x^{{1}\over{\frac{1}{2}}}=x^2?????\]

Answer 3

Lol, very awkward but I would say yes, all that is just x^{2}

Answer 4

\(\Huge \sqrt[1/2]{4}= 16 ?\)

Answer 5

have to think thru lol

Answer 6

I was considering it, lol, but yeah, basically I was actually trying to find a rule for the derivative of a function \[\Huge \sqrt[n]{x}\]

Answer 7

\[\Huge \frac{ d }{ dx }\sqrt[n]{x}=\frac{ 1 }{ n~\times~\sqrt[n]{x^{n-1}}}\]

Answer 8

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