If x^m yⁿ = (x + y)^{m + n}, prove that 

We are given,
If x^m yⁿ = (x + y)^{m + n}

Taking log on both sides, we get
log x^m yⁿ = log(x + y)^{m + n}
log x^m + log yⁿ = m+n log(x + y)
m logx + n logy = m+n log (x+y)

Differentiating above equation w.r.t. x, we get