Question

If u=sin^-2 (y/x) then du/dX=​

Answer 1

Answer:

$\frac{du}{dx} = - 2 \frac{y}{ {x}^{2} } \ { \sin}^{ - 3} ( \frac{y}{x} ) \cos( \frac{y}{x} )$

Step-by-step explanation:

$if \: u = { \sin }^{ - 2} ( \frac{y}{x} )$

$\frac{du}{dx} = - 2 { \sin}^{ - 3} ( \frac{y}{x} ) { \cos } ( \frac{y}{x} )\frac{y}{ {x}^{2} }$

$\frac{du}{dx} = - \frac{y}{ {x}^{2} } 2 {\sin }^{ - 3} ( \frac{y}{x}) {\cos} (\frac{y}{x} )$

we can solve another way

for second method,

we know that ,

${ \sin}^{ - 2} ( \frac{y}{x} ) = {cosec}^{2} ( \frac{y}{x} )$

then,

$\frac{du}{dx} = 2cosec( \frac{y}{x} ) (- cosec( \frac{y }{x} ).cot( \frac{y}{x} ))\frac{y}{ {x}^{2} }$

$\frac{du}{dx} = - \frac{ {y} }{ {x}^{2} } 2 {cosec}^{3} ( \frac{y}{x} ) \cos( \frac{y}{x} )$