If xm.yn=(x+y)m+n, then dydx is
yx
x+yxy
xy
xm.yn=(x+y)m+n ⇒m ln(x)+n ln(y)=(m+n) ln(x+y) Differentiating both sides. ∴mx+nydydx=m+nx+y(1+dydx) ⇒(mx−m+nx+y)=(m+nx+y−ny)dydx ⇒my−nxx(x+y)=(my−nxy(x+y))dydx⇒dydx=yx
If xm.yn=(x+y)m+n,thendydx=