Question

$d\left({\tan }^{-1}\frac{y}{x}\right)=$.

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

Answer 1

The correct option is B $\frac{x\mathrm{dy}-y\mathrm{dx}}{x^2+y^2}$

$\mathrm{Derivative}\mathrm{of}{\tan }^{-1}\left(x\right)=\frac{1}{1+x^2}\mathrm{Applying}\mathrm{chain}\mathrm{rule}:d\left({\tan }^{-1}\frac{y}{x}\right)=\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\times d\left(\frac{y}{x}\right)=\frac{1}{1+{\left(\frac{y}{x}\right)}^2}\times \frac{x\mathrm{dy}-y\mathrm{dx}}{x^2}=\frac{x\mathrm{dy}-y\mathrm{dx}}{x^2+y^2}$