If (x+y)^(mn) = x^my^n then show y' = (m y(x + y - n x)) / (nx(m y - x - y ))?

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3 Answers
May 20, 2017

See below.

Explanation:

If f(x,y)=(x+y)^(nm)-x^m y^n=0 then

f_x dx+f_y dy = 0 so

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

now making the substitution

(x+y)^(nm) = x^my^n after some minor simplifications we have

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

if instead f(x,y)=(x+y)^(m+n)-x^my^n=0

the answer would be much simpler

dy/dx = y/x

May 20, 2017

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: y' = (my(x+y-nx))/(nx(ym-x-y))

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 x+y

mn + mny' = m*(x+y)/x + n*(y'(x+y))/y
mn + mny' = m + (my)/x + ny' + (nxy')/y

Group all of the y' terms together on one side.

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 (yx)/(yx)

y' = (my(x+y-nx))/(nx(my-x-y))

Final Answer

May 20, 2017

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 (x+y)xy :

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 (x+y)^(mn) = x^my^n

mn x^m y^n xy (1+y')= (x+y)x^m y^n(n x y' + m y )

Cancel x^m y^n:

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 y' terms:

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