Question

If `"x"^"m"*"y"^"n" = ("x + y")^("m + n")`, then `"dy"/"dx" = "______"/"x"`

Answer 1

If `"x"^"m"*"y"^"n" = ("x + y")^("m + n")`, then `"dy"/"dx" = bbunderline"y"/"x"`

Explanation:

x^m.yⁿ = `("x + y")^("m + n")`

log(x^m.yⁿ) = log`("x + y")^("m + n")`

mlogx + nlogy = (m + n) log(x + y)      ...`[(log(ab)=loga+logb),(logm^n=nlogm)]`

Diff. w.r.t.x.

`mxx1/x+nxx1/y.dy/dx=(m+n)1/(x+y)xx(1+dy/dx)` 

`m/x+n/y.dy/dx=(m+n)/(x+y)+(m+n)/(x+y)dy/dx`

`n/ydy/dx-(m+n)/(x+y)dy/dx=(m+n)/(x+y)-m/x`

`dy/dx(n/y-(m+n)/(x+y))=(m+n)/(x+y)-m/x`

`dy/dx[(nx+ny-my-ny)/(y(x+y))]=(mx+mx-mx-my)/((x+y)x)`

`dy/dx=(nx-my)/x xxy/(nx-my)`

`dy/dx=y/x`