Question

If $u={\sin }^{-1}\left(\frac{y}{x}\right),$ then $\frac{\partial u}{\partial x}$ is equal to

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

Answer 1

The correct option is D

$\frac{-y}{x\surd (x^2–y^2)}$

Explanation for the correct answer:

Finding the value of $\frac{\partial u}{\partial x}$:

Given,

$\begin{array}{rcl}u& =& {\sin }^{-1}\left(\frac{y}{x}\right)\end{array}$

Differentiate partially the above equation with respect to $x$.

$\begin{array}{rcl}\frac{\partial u}{\partial x}& =& \left(\frac{1}{\surd \left(1-{\left({\displaystyle \frac{y}{x}}\right)}^2\right)}\right)\times \left(\frac{-y}{x^2}\right)\\ & =& \left(\frac{x}{\surd (x^2–y^2)}\right)\times \left(\frac{-y}{x^2}\right)\\ & =& \left(\frac{-y}{\surd (x^2–y^2)}\right)\end{array}$

Hence, the correct option is D.