Question

derivative of tan inverse y/x with respect to x​

Answer 1

Answer:

$\bf\frac{\partial{u}}{\partial{x}}=\frac{-y}{x^2+y^2}$

Step-by-step explanation:

$Let\:u=tan^{-1}\frac{y}{x}$

$\text{Then,}$

$\frac{\partial{u}}{\partial{x}}=\frac{1}{1+(\frac{y}{x})^2}\frac{\partial(\frac{y}{x})}{\partial{x}}$

$\frac{\partial{u}}{\partial{x}}=\frac{1}{1+\frac{y^2}{x^2}}\frac{\partial(yx^{-1})}{\partial{x}}$

$\frac{\partial{u}}{\partial{x}}=\frac{1}{\frac{x^2+y^2}{x^2}}(-yx^{-2})$

$\frac{\partial{u}}{\partial{x}}=\frac{x^2}{x^2+y^2}(\frac{-y}{x^2})$

$\frac{\partial{u}}{\partial{x}}=\frac{1}{x^2+y^2}(-y)$

$\implies\boxed{\bf\frac{\partial{u}}{\partial{x}}=\frac{-y}{x^2+y^2}}$