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AIEEEAIEEE 2006Continuity and Differentiability
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Solution:
∵xm⋅ym=(x+y)m+n
Taking log on both sides, we get mlogx+nlogy=(m+n)log(x+y)
On differentiating with respect to x, we get xm+yndxdy=(x+y)(m+n)(1+dxdy) ⇒dxdy(x+ym+n−yn)=xm−x+ym+n ⇒dxdy(y(x+y)my+ny−nx−ny) =x(x+y)mx+my−mx−nx ⇒dxdy=xy