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There's need to find y' of:

$$\arctan(y/x)=\ln\sqrt{x^2 + y^2}$$

Tried:

$\dfrac{1}{(1+(y/x)^2)}*(\dfrac{y}{x})'=(x^2+y^2)^\dfrac{-1}{2}*(\dfrac{1}{2})*(x^2+y^22)^\dfrac{-1}{2} * (2x+2y'*y')$

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  • $\begingroup$ Please, rewrite the problem ! $\endgroup$ Nov 23, 2013 at 16:38
  • $\begingroup$ Take derivative from both sides and solve for $y'$. $\endgroup$
    – hhsaffar
    Nov 23, 2013 at 16:42

3 Answers 3

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$$\arctan(\frac{y}{x})=\ln\sqrt{x^2+y^2}$$ $$\frac{1}{1+(y/x)^2}\frac{y'x-y}{x^2}=\frac{x+yy'}{\sqrt{x^2+y^2}}$$ solve for $y'$

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$$\arctan \frac yx=\ln\sqrt{x^2+y^2}=\frac12\ln(x^2+y^2)$$

Using Chain Rule, $$\frac{d \left(\arctan \frac yx\right)}{d\left(\frac yx\right) }\cdot \frac{d\left(\frac yx\right) }{dx}=\frac12\frac{d\left(\ln(x^2+y^2)\right)}{d(x^2+y^2)}\cdot\frac{d(x^2+y^2)}{dx}$$

$$\frac1{1+\frac{y^2}{x^2}}\frac{x\frac{dy}{dx}-y}{x^2}=\frac1{2(x^2+y^2)}\left(2x+2y\frac{dy}{dx}\right)$$

Now simplify and take out $\frac{dy}{dx}$

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You can use the Implicit function theorem

Put $F(x,y)=\arctan(y/x)-\ln\sqrt{x^2 + y^2}$

According to the Implicit function theorem we have :

$y'=-\frac{F_{y}^{'}(x,y)} {F_{x}^{'}(x,y)}=\frac{\frac{x+y}{x^2+y^2}}{\frac{x-y}{x^2+y^2}}=\frac{x+y}{x-y}$

Hope it will held you.

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