Question

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!)

Answer 1

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`

Answer 2

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

Answer 3

`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