If xmyn=(x+y)m+n, thendydxis equal to

Given that, xmyn=(x+y)m+nTaking on both sides, =we getmlog⁡x+nlog⁡y=(m+n)log⁡(x+y)On differentiating w.r.t. x, we getmx+nydydx=(m+n)(x+y)1+dydx⇒ dydxm+nx+y−ny=mx−m+nx+y⇒dydxmy+ny−nx−nyy(x+y)=mx+my−mx−nxx(x+y)∴dydx=yx