if (X^n)(Y^m) = (x+y)^(m+n), can you prove that (dy/dx = y/x)?

Question

if (X^n)(Y^m) = (x+y)^(m+n), can you prove that (dy/dx = y/x)

thnks

Answer 1

x^ny^m = (x+y)^(m+n)

differentiate both sides

and use the formula d/dx(uv) = u dv/dx + vdu/dx
d/dx(x^ny^m) = x^n(d/dx)y^m + y^m d/dx(x^n)
= x^n (m)y^(m-1) dy/dx + y^mnx^(n-1)
= x^(n-1) y(m-1)(m xdy/dx + ny)

d/dx(x+y)^(m+n) = (m+n)(1+ dy/dx)(x+y)^(m+n-1)

from above 2

x^(n-1)y(m-1) (m xdy/dx + ny ) = (m+n)(x+y)^(m+n-1)
or multiply by xy on both sides

x^ny^m(mdy/dx+ny) = (m+n) xy(x+y)^(m+n-1)(1+dy/dx)
or (x+y)^(n+m)). (mdy/dx +ny) = (m+n) xy(x+y)^(m+n-1)(1+dy/dx)
or (x+y) (mdy/dx+ny) = (m+n) xy(1+dy/dx)

I am unable to proceed but some one may

Answer 2

This problem has a lot of blind alleys, but one way to prove this is to consider a rational fraction of the form: a/(an-b) where n > a > b are integers. Subtracting 1/n from it, we have: a/(an-b) - 1/n = (an - an + b)/(n(an-b)) = b/(an² - bn) where we replace an² - bn with bn' - c, where b > c. Thus, we have a remainder fraction where b < a, and this process can be repeated until we're left with a fraction of the form 1/n'', and the algorithmic process ends with a finite sum of fractions. It should be noted that for any given rational fraction 0 < x < 1, there is a multiplicity of ways it can be expressed in the form 1/a + 1/b + 1/c + ..., where a, b, c... are all distinct integers. This problem reminds me of Turing's Halting Problem, in that what we're being asked to do is to show that if we subtract fractions of the form 1/n successively from the original rational fraction, the process will halt eventually. It's kind of fun to try doing it using random fractions, eventually it halt with sometimes some pretty far out fractions. Addendum: If for any reason during the algorithmic process we come up with 2 fractions that are alike, we can use amir11elad's method of replacing one of the fractions with two others, and if that causes another match, we can repeat, until after a finite number of step there will be no further matches and all the fractions would be distinct. Addendum 2: Here's the proof again, more concisely: 1) Subtract fractions of the form 1/n from the rational number 0 < x < 1 successively (with distinct n) until we are left with a fraction of the form a/(an-b), where n > a > b are integers. 2) Subtract 1/n from this fraction, so that we are left with a fraction of the form b/(bn-c), where n > b > c are integers. 3) Continue until we're left with the fraction of the form 1/n. If you have a series of integers a > b > c > ...., eventually, you will end up with 1. That's why the series will always be finite. Addendum 3: Hear ye, Alexey V

Answer 3

yeah sure since x-y/2= e*6 then f=98 from there you understand that dy=h and dx=s then i dont know....im a bit lost i ll get back to you for more in a while give me a minute!

Answer 4

If dy/dx = y/x, so y = x

=>
(x^n)(y^m) = (x^n)(x^m) = x^(m+n)

and
(x+y)^(m+n) = (x+x)^(m+n) = (2x)^(m+n)

=>
x^(m+n) <> (2x)^(m+n)

x^(m+n) is not (2x)^(m+n), only if x=0.

₢