Question

If $u = {\sin ^{ - 1}}\left( {\dfrac{y}{x}} \right)$, then find $\dfrac{{\partial u}}{{\partial x}}$.
A. $ - \dfrac{y}{{{x^2} + {y^2}}}$
B. $\dfrac{x}{{\sqrt {1 - {y^2}} }}$
C. $ - \dfrac{y}{{\sqrt {{x^2} - {y^2}} }}$
D. $ - \dfrac{y}{{x\sqrt {{x^2} - {y^2}} }}$

Answer 1

Hint: To find the partial derivative of the given equation we will find the derivative with respect to $x$. In the derivative, we will consider $y$ as a constant.

Formula Used:
$\dfrac{\partial }{{\partial x}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}$
$\dfrac{{\partial u}}{{\partial x}}\left( {\dfrac{1}{x}} \right) = - \dfrac{1}{{{x^2}}}$

Complete step by step solution:
Given equation is $u = {\sin ^{ - 1}}\left( {\dfrac{y}{x}} \right)$
Now find the partial derivative of $u = {\sin ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ with respect to $x$.
Apply chain rule:
$\dfrac{\partial }{{\partial x}}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{y}{x}} \right)} \right] = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{y}{x}} \right)}^2}} }}\dfrac{\partial }{{\partial x}}\left( {\dfrac{y}{x}} \right)$
$ \Rightarrow \dfrac{\partial }{{\partial x}}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{y}{x}} \right)} \right] = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{y}{x}} \right)}^2}} }} \cdot y\dfrac{\partial }{{\partial x}}\left( {\dfrac{1}{x}} \right)$

$ \Rightarrow \dfrac{\partial }{{\partial x}}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{y}{x}} \right)} \right] = \dfrac{1}{{\sqrt {1 - {{\left( {\dfrac{y}{x}} \right)}^2}} }} \cdot y\left( { - \dfrac{1}{{{x^2}}}} \right)$
Simplify the above equation

$ \Rightarrow \dfrac{\partial }{{\partial x}}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{y}{x}} \right)} \right] = \dfrac{1}{{\sqrt {\dfrac{{{x^2} - {y^2}}}{{{x^2}}}} }} \cdot \left( { - \dfrac{y}{{{x^2}}}} \right)$
$ \Rightarrow \dfrac{\partial }{{\partial x}}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{y}{x}} \right)} \right] = \dfrac{x}{{\sqrt {{x^2} - {y^2}} }} \cdot \left( { - \dfrac{y}{{{x^2}}}} \right)$
Cancel out $x$ from the denominator and numerator of the right-side expression
$ \Rightarrow \dfrac{\partial }{{\partial x}}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{y}{x}} \right)} \right] = - \dfrac{y}{{x\sqrt {{x^2} - {y^2}} }}$

Option ‘D’ is correct

Note: Partial derivative is almost the same as normal derivative. But in the partial derivative we are considering only one variable and other variables are treated as constant. When we find the partial derivative $\dfrac{\partial }{{\partial x}}$, then we consider $x$ as a variable and $y$ as a constant.