Respuesta :

LammettHash
[tex](x+y)^{m+n}=x^my^n[/tex]
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[(x+y)^{m+n}\right]=\dfrac{\mathrm d}{\mathrm dx}\left[x^my^n\right][/tex]

By the power/chain/product rules,

[tex](m+n)(x+y)^{m+n-1}\dfrac{\mathrm d}{\mathrm dx}[x+y]=\dfrac{\mathrm d}{\mathrm dx}\left[x^m\right]y^n+x^m\dfrac{\mathrm d}{\mathrm dx}\left[y^n\right][/tex]
[tex](m+n)(x+y)^{m+n-1}\left(1+\dfrac{\mathrm dy}{\mathrm dx}\right)=mx^{m-1}y^n+nx^my^{n-1}\dfrac{\mathrm dy}{\mathrm dx}[/tex]
[tex]\left((m+n)(x+y)^{m+n-1}-nx^my^{n-1}\right)\dfrac{\mathrm dy}{\mathrm dx}=mx^{m-1}y^n-(m+n)(x+y)^{m+n-1}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{mx^{m-1}y^n-(m+n)(x+y)^{m+n-1}}{(m+n)(x+y)^{m+n-1}-nx^my^{n-1}}[/tex]

Otras preguntas