Question

(x+y)^(m+n) = x^m.y^n find dy/dx without using log

Answer 1

`(x+y)^(m+n)=x^my^n`

`(\mathrm d)/(\mathrm dx)\left[(x+y)^(m+n)\right]=(\mathrm d)/(\mathrm dx)\left[x^my^n\right]`

By the power/chain/product rules,


`(m+n)(x+y)^(m+n-1)(\mathrm d)/(\mathrm dx)[x+y]=(\mathrm d)/(\mathrm dx)\left[x^m\right]y^n+x^m(\mathrm d)/(\mathrm dx)\left[y^n\right]`

`(m+n)(x+y)^(m+n-1)\left(1+(\mathrm dy)/(\mathrm dx)\right)=mx^(m-1)y^n+nx^my^(n-1)(\mathrm dy)/(\mathrm dx)`

`\left((m+n)(x+y)^(m+n-1)-nx^my^(n-1)\right)(\mathrm dy)/(\mathrm dx)=mx^(m-1)y^n-(m+n)(x+y)^(m+n-1)`

`(\mathrm dy)/(\mathrm dx)=(mx^(m-1)y^n-(m+n)(x+y)^(m+n-1))/((m+n)(x+y)^(m+n-1)-nx^my^(n-1))`