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!)

3 Answers
May 20, 2017

Answer:

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

Answer:

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

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 # (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