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Question
If `"x"^"m"*"y"^"n" = ("x + y")^("m + n")`, then `"dy"/"dx" = "______"/"x"`
Solution
If `"x"^"m"*"y"^"n" = ("x + y")^("m + n")`, then `"dy"/"dx" = bbunderline"y"/"x"`
Explanation:
xm.yn = `("x + y")^("m + n")`
log(xm.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`
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