यदि (If) `(x+y)^(m+n)=x^(m)y^(m)`, दिखाएँ कि (show that) `(dy)/(dx)=(y)/(x)`

Explanation

दिया है ,`(x+y)^(m+n)=x^(m)y^(m)` … (1)
दोनों तरफ x के सापेक्ष अवकलित (Differentiate) करने पर हमें मिलता है,
`(d)/(dx)(x+y)^(m+n)=(d)/(dx)(x^(m))*y^(n)+x^(m)*(d)/(dx)(y^(n))`
या `(d)/(d(x+y))(x+y)^(m+n)*(d)/(dx)(x+y)=mx^(m-1)*y^(n)+x^(m)(d)/(dy)(y^(n))*(dy)/(dx)`
या `(m+n)(x+y)^(m+n-1)*(1+(dy)/(dx))=mx^(m-1)*y^(n)+x^(m)*ny^(n-1)(dy)/(dx)`
या `(m+n)((x+y)^(m+n))/((x+y)^(1))(1+(dy)/(dx))=mx^(m-1)*y^(n)+nx^(m)*y^(n-1)(dy)/(dx)`
या `(m+n)(x^(m)y^(n))/(x+y)(1+(dy)/(dx))=x^(m-1)y^(n-1)(my+nx(dy)/(dx)) " " [ because(1), (x+y)^(m+n)=x^(m)y^(n)]`
या `(m+n)/(x+y)*xy(1+(dy)/(dx))=my+nx(dy)/(dx)`
या `(m+n)xy+(m+n)xy(dy)/(dx)=(x+y)my+(x+y)nx(dy)/(dx)`
या `[(m+n)xy-(x+y)nx](dy)/(dx)=(x+y)my-(m+n)xy`
या `x(my+ny-nx-ny)(dy)/(dx)=y(mx+my-nx)`
या `x(my-nx)(dy)/(dx)=y(my-nx)therefore(dy)/(dx)=(y)/(x)`