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Question

If xmyn=(x+y)m+n. Prove that d2ydx2=0.

Solution
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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(nym+nx+y)=m+nx+ymx
dydx(n(x+y)y(m+n)y(x+y))=x(m+n)m(x+y)x(x+y)

On simplifying we get,

dydx(nxmyy)=nxmyx
dydx=yx

Differentiating on both sides w.r.t x we get,
Apply quotient rule

d2ydx2=x.dydxy.1x2
x.dydxyx2
This can be written as
1xdydxyx2

But dydx=yx

So substituting this we get,
yx2yx2=0
Hence proved

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