If ` x^(p) + y^(q) = (x + y)^(p+q) , " then" (dy)/(dx)` is

Options

A. ` - ( x ) /( y ) `
B. ` ( x ) /( y ) `
C. `- ( y ) / ( x ) `
D. ` ( y ) /( x ) `

Answer

D

Explanation

if ` x ^p + y ^ q = ( x + y ) ^ ( p + q ) `
Taking log on both sides ,
` p log x + q log y = ( p + q ) log ( x + y ) `
On differentiating w.r.t x, we get
`( p ) /( x ) + ( q ) /( y ) * ( dy ) /( dx ) = (( p + q ))/( ( x + y)) ( 1 + (dy ) /(dx )) `
` { (p) / ( x ) - ( p+ q )/( x + y )} = { ( p + q )/( x + y ) - ( q )/( y ) } (dy ) /( dx ) `
` { ( px + py - px - qx ) /( x ( x + y ))} = { ( py + qy - qx - qy ) /( y ( x + y ))} (dy)/(dx ) `
` rArr (( py - qx ) ) /( x ) = ((py- qx ) ) / (y ) * ( d y ) / ( dx ) `
`rArr ( dy)/ (dx ) = (y )/ ( x ) `