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- Thread starter redshift
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redshift said:

You know that

[tex] [\arctan(f(x))]'=\frac{f'(x)}{1+f^2 (x)} [/tex]

,so apply the formula correctly.This is of course if "y" and "x" are independent variables.

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HallsofIvy

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by the chain rule.

Now multiply both numerator and denominator by x

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HallsofIvy said:

by the chain rule.

Now multiply both numerator and denominator by x^{2}.

Apparently the sofware didn't read the paranthesis you've written,so it should be:

[tex]z'= \frac{1}{1+(\frac{y}{x})^2} (-\frac{y}{x^2}) =-\frac{y}{x^2+y^2}[/tex]

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