If xmyn=(x+y)m+n. Prove that d2ydx2=0.
Given: xmyn=(x+y)m+n
Take log on both side
logxm+logyn+(m+n)log=y(x+y)
mlogx+nlogy=(m+n)log(x+y)
Differentiating w.r.t x on both sides we get,
m.1x+n1ydydx=(m+n).1(x+y)(1+dydx)
mx+nydydx=m+nx+y(1+dydx)
dydx(ny−m+nx+y)=m+nx+y−mx
dydx(n(x+y)−y(m+n)y(x+y))=x(m+n)−m(x+y)x(x+y)
On simplifying we get,
dydx(nx−myy)=nx−myx
∴dydx=yx
Differentiating on both sides w.r.t x we get,
Apply quotient rule
d2ydx2=x.dydx−y.1x2
⇒x.dydx−yx2
This can be written as
1xdydx−yx2
But dydx=yx
So substituting this we get,
yx2−yx2=0
Hence proved