Question

If $u=\tan ({\tan }^{-1}x+{\tan }^{-1}y)$

then $\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}=?$

• A. $\frac{x^2+y^2}{(1-xy{)}^2}$
• B. $\frac{x^2-y^2}{(1-xy{)}^2}$
• C. $u$
• D. $\frac{y^2-x^2}{(1-xy{)}^2}$

Answer 1

Answer: (D)

$\frac{y^2-x^2}{(1-xy{)}^2}$

$u=\frac{x+y}{1-xy}use\left(tan\right(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B})$

$\frac{du}{dx}=\frac{(1-xy)+(x+y)y}{(1-xy{)}^2}=\frac{1+y^2}{(1-xy{)}^2}$

$\frac{du}{dy}=\frac{(1-xy)+(x+y)(-x)}{(1-xy{)}^2}=\frac{1+y^2}{(1-xy{)}^2}$

$\frac{du}{dx}-\frac{du}{dy}=\frac{y^2-x^2}{(1-xy{)}^2}$