If #x^m * y^n = (x + y)^(m + n)#. Then how will you prove that #dy/dx = y/x#??
1 Answer
This is very similar to this problem:
https://socratic.org/questions/what-is-the-derivative-of-x-y-mn-x-m-y-n
Here, we have:
# (x+y)^(m+n) = x^my^n #
Taking Natural logarithms of both sides:
# ln {(x+y)^(m+n)} = ln{ x^my^n} #
And using the rules of logs we can manipulate as follows:
# (m+n)ln (x+y) = ln x^m + lny^n #
# :. (m+n)ln (x+y) = mln x + nlny #
Differentiate Implicitly:
# (m+n) * 1/(x+y) * (1+y') = m/x + n/y * y' #
# :. (m+n) (1+y')/(x+y) = (my + nxy')/(xy) #
Cross multiply by
# (m+n) (1+y')xy = (my + nxy')(x+y) #
Multiply out:
# mxy+mxyy' + nxy+nxyy' = mxy + nx^2y' + my^2 + nxyy' #
# :. mxyy' + nxy = nx^2y' + my^2 #
Finally, Collect term,s and tidy up:
# mxyy' -nx^2y' = my^2 - nxy#
# :. (my -nx)xy' = (my - nx)y#
# :. xy' = y#
# :. y' = y/x# QED
This assumes that where applicable the denominators are non-zero to permit the division to occur.
