Question

If $\tan^{-1} \left(\frac{x}{y}\right) + \log \sqrt{x^{2} +y^{2}} = 0$ , then $\frac{dx}{dy} =$

• A. $\frac{x^{2} +y^{2}}{x^{2} -y^{2}}$
• B. $\frac{x - y}{x + y}$
• C. $\frac{x + y}{x - y}$
• D. $\frac{x^{2} -y^{2}}{x^{2} +y^{2}}$

Answer 1

Answer: (B)

$\frac{x - y}{x + y}$

Answer 2

$\tan^{-1} \left(\frac{x}{y}\right) + \log\sqrt{x^{2} +y^{2}} = 0$
Differentiating w.r.t. 'y', we get
$\left(\frac{1}{1+\left(\frac{x}{y}\right)^{2}}\right)\left(\frac{y \frac{dx}{dy} -x.1}{y^{2}}\right) + \frac{1}{\sqrt{x^{2}+y^{2}}} . \frac{1}{2} \frac{\left(2x \frac{dx}{dy} +2y\right)}{\sqrt{x^{2} +y^{2}}} = 0$
$\Rightarrow \left(\frac{y^{2}}{x^{2} +y^{2}}\right)\left(\frac{y \frac{dx}{dy}-x}{y^{2}}\right) + \frac{x \frac{dx}{dy }+y}{\left(x^{2} +y^{2}\right)} = 0$
$\Rightarrow \frac{y \frac{dx}{dy} -x+x \frac{dx}{dy} +y}{x^{2} +y^{2}} = 0$
$\Rightarrow \left(y+x\right) \frac{dx}{dy} +y -x = 0 \Rightarrow \frac{dx}{dy} = \frac{x-y}{x+y}$