Question

if u=sin inverse x/y + tan inverse y/x .show that xdu/dx + y du/dy =0​

Answer 1

Answer:

$\bf\,x\,\frac{\partial{u}}{\partial{x}}+y\,\frac{\partial{u}}{\partial{y}}=0$

Step-by-step explanation:

Given:

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

Differentiate "u" partially with respect to x

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

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

$\implies\frac{\partial{u}}{\partial{x}}=\frac{1}{\sqrt{y^2-x^2}}-\frac{y}{x^2+y^2}$

Differentiate "u" partially with respect to y

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

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

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

Now,

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

$\implies\,x\,\frac{\partial{u}}{\partial{x}}+y\,\frac{\partial{u}}{\partial{y}}=\frac{x}{\sqrt{y^2-x^2}}-\frac{xy}{x^2+y^2}-\frac{x}{\sqrt{y^2-x^2}}+\frac{xy}{x^2+y^2})$

$\implies\,\boxed{x\,\frac{\partial{u}}{\partial{x}}+y\,\frac{\partial{u}}{\partial{y}}=0}$