I am trying to see right now how these became equivalent:
$\displaystyle \frac{1}{2}((x-a)^2+y_0^2)^-^1^/^2(2(x-a))=\frac{x-a}{\sqrt(x-a)^2+y_0^2}$
Thanks in advance...
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I am trying to see right now how these became equivalent:
$\displaystyle \frac{1}{2}((x-a)^2+y_0^2)^-^1^/^2(2(x-a))=\frac{x-a}{\sqrt(x-a)^2+y_0^2}$
Thanks in advance...
$\displaystyle x^{-k} = \dfrac{1}{x^k}$ and $\displaystyle x^{1/n} = \sqrt[n]{x}$ (where $\displaystyle k$ is any real number and $\displaystyle n$ is a positive integer). Do you see it now?
$\displaystyle \frac{1}{2}*\frac{2(x-a)}{((x-a)^2+y_0^2)^1^/^2}$?
I think this is maybe it. Thank you for your reply.
I presume you know that $\displaystyle \frac{1}{2}(2)= 1$. You should also know that $\displaystyle A^{1/2}= \sqrt{A}$ and $\displaystyle A^{-1/2}= \frac{1}{A^{1/2}}= \frac{1}{\sqrt{A}}$Quote:
Originally Posted by sepoto
