If (x+y)^(mn) = x^my^n then show y' = (m y(x + y - n x)) / (nx(m y - x - y ))?
(portions of this question have been edited or deleted!)
(portions of this question have been edited or deleted!)
3 Answers
See below.
Explanation:
If
now making the substitution
if instead
the answer would be much simpler
Use implicit differentiation with natural logs.
The answer I got was slightly different, so you may want to check to see if you mistyped the answer.
My answer:
Explanation:
Take the natural log of both sides and differentiate from there.
ln((x+y)^(mn)) = ln(x^my^n)
color(white)"-"mn ln(x+y) = m ln(x) + n ln(y)
Differentiating gives:
mn*(1+y')/(x+y) = m*1/x + n*(y')/y
Now multiply both sides by
mn + mny' = m*(x+y)/x + n*(y'(x+y))/y
mn + mny' = m + (my)/x + ny' + (nxy')/y
Group all of the
mny'-(nxy')/y-ny' = m+(my)/x-mn
y'(mn-(nx)/y-n) = m+(my)/x-mn
Now divide and simplify.
y' = (m+(my)/x-mn)/(mn-(nx)/y-n)
y' = (m(1+y/x-n))/(n(m-x/y-1))
Multiply by
y' = (my(x+y-nx))/(nx(my-x-y))
Final Answer
y' = (m y(x + y - n x)) / (nx(m y - x - y ))
Explanation:
We have:
(x+y)^(mn) = x^my^n
If we differentiate implicitly and apply the product rule:
(mn)(x+y)^(mn-1) d/dx(x+y )= (x^m)(d/dxy^n) + (d/dxx^m)(y^n)
:. (mn)(x+y)^(mn-1) (1+y')= (x^m)(ny^(n-1)y') + (mx^(m-1))(y^n)
:. mn(x+y)^(mn-1) (1+y')= nx^my^(n-1)y' + mx^(m-1)y^n
Multiply By
mn(x+y)^(mn)xy (1+y')= (x+y)(n x x^m y^n y' + m x^m y y^n)
On the LHS, replace
mn x^m y^n xy (1+y')= (x+y)x^m y^n(n x y' + m y )
Cancel
mn xy (1+y')= (x+y)(n x y' + m y )
:. mn xy +mn xyy' = n x^2 y' + mx y + n x y y' + m y^2
Collect
mn xyy'-n x^2 y'-n x y y' = + mx y + m y^2 - mn xy
:. nx(m y - x - y )y' = m y(x + y - n x)
Leading to:
y' = (m y(x + y - n x)) / (nx(m y - x - y ))
And we conclude that the given solution is incorrect