Need hint on 'simple' differentiation problem

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To differentiate z = arctan(y/x) with respect to x, the correct application of the chain rule is crucial. The derivative is derived as z' = (1/(1+(y/x)^2)) * (-y/x^2). After simplifying, the final result is z' = -y/(x^2+y^2). The initial confusion stemmed from misapplying the derivative formula for arctan. Properly applying the chain rule clarifies the differentiation process.
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I need to differentiate z = artan(y/x) with respect to x. Somehow z' = y/(1 +(1/x)^2) doesn't seem right.
 
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The derivative of arctan x = 1/(1 + x^2). There's a proof of that here.
 
redshift said:
I need to differentiate z = artan(y/x) with respect to x. Somehow z' = y/(1 +(1/x)^2) doesn't seem right.

You know that
[\arctan(f(x))]'=\frac{f'(x)}{1+f^2 (x)}
,so apply the formula correctly.This is of course if "y" and "x" are independent variables.
 
z'= \frac{1}{1+(\frac{y}{x})^2} \(-\frac{y}{x^2}\)
by the chain rule.

Now multiply both numerator and denominator by x2.
 
HallsofIvy said:
z'= \frac{1}{1+(\frac{y}{x})^2} \(-\frac{y}{x^2}\)
by the chain rule.

Now multiply both numerator and denominator by x2.

Apparently the sofware didn't read the paranthesis you've written,so it should be:
z'= \frac{1}{1+(\frac{y}{x})^2} (-\frac{y}{x^2}) =-\frac{y}{x^2+y^2}
 
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