1.

If `x^(a)cdoty^(b)=(x+y)^(a+b)` then prove that `(dy)/(dx)=y/x`

Answer» `x^(a)cdoty^(b)=(x+y)^(a+b)`
`rArrlog(x^(a)cdoty^(b))=log(x+y)^(a+b)`
`rArrlogx^(a)+logy^(b)=(a+b)log(x+y)`
`rArralogx+blogy=(a+b)log(x+y)`
Differentiate both sides with respect to x
`a/x+b/y(dy)/(dx)=((a+b))/((x+y))cdotd/(dx)(x+y)`
`=(a+b)/(x+y)(1+(dy)/(dx))`
`=(a+b)/(x+y)+(a+b)/(x+y)+(a+b)/(x+y)cdot(dy)/(dx)`
`rArr(b/y-(a+b)/(x+y))(dy)/(dx)=(a+b)/(x+y)-a/x`
`rArr(b(x+y)-y(a+b))/(y(x+y))cdot(dy)/(dx)-(x(a+b)-a(x+y))/(x(x+y))`
`rArr(bx+by-ay-by)/ycdot(dy)/(dx)=(ax+bx-ax-ay)/x`
`rArr(bx-ay)/ycdot(dy)/(dx)=(bx-ay)/xrArr(dy)/(dx)=y/x`
Hence proved.


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