# 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!)

May 20, 2017

See below.

#### Explanation:

If $f \left(x , y\right) = {\left(x + y\right)}^{n m} - {x}^{m} {y}^{n} = 0$ then

${f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} = 0$ 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

${\left(x + y\right)}^{n m} = {x}^{m} {y}^{n}$ after some minor simplifications we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{m \left(\left(n - 1\right) x - y\right) y}{n x \left(x + y - m y\right)}$

if instead $f \left(x , y\right) = {\left(x + y\right)}^{m + n} - {x}^{m} {y}^{n} = 0$

the answer would be much simpler

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x}$

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: $y ' = \frac{m y \left(x + y - n x\right)}{n x \left(y m - x - y\right)}$

#### Explanation:

Take the natural log of both sides and differentiate from there.

$\ln \left({\left(x + y\right)}^{m n}\right) = \ln \left({x}^{m} {y}^{n}\right)$
$\textcolor{w h i t e}{\text{-}} m n \ln \left(x + y\right) = m \ln \left(x\right) + n \ln \left(y\right)$

Differentiating gives:

$m n \cdot \frac{1 + y '}{x + y} = m \cdot \frac{1}{x} + n \cdot \frac{y '}{y}$

Now multiply both sides by $x + y$

$m n + m n y ' = m \cdot \frac{x + y}{x} + n \cdot \frac{y ' \left(x + y\right)}{y}$
$m n + m n y ' = m + \frac{m y}{x} + n y ' + \frac{n x y '}{y}$

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

$m n y ' - \frac{n x y '}{y} - n y ' = m + \frac{m y}{x} - m n$
$y ' \left(m n - \frac{n x}{y} - n\right) = m + \frac{m y}{x} - m n$

Now divide and simplify.

$y ' = \frac{m + \frac{m y}{x} - m n}{m n - \frac{n x}{y} - n}$

$y ' = \frac{m \left(1 + \frac{y}{x} - n\right)}{n \left(m - \frac{x}{y} - 1\right)}$

Multiply by $\frac{y x}{y x}$

$y ' = \frac{m y \left(x + y - n x\right)}{n x \left(m y - x - y\right)}$

May 20, 2017

$y ' = \frac{m y \left(x + y - n x\right)}{n x \left(m y - x - y\right)}$

#### Explanation:

We have:

${\left(x + y\right)}^{m n} = {x}^{m} {y}^{n}$

If we differentiate implicitly and apply the product rule:

$\left(m n\right) {\left(x + y\right)}^{m n - 1} \frac{d}{\mathrm{dx}} \left(x + y\right) = \left({x}^{m}\right) \left(\frac{d}{\mathrm{dx}} {y}^{n}\right) + \left(\frac{d}{\mathrm{dx}} {x}^{m}\right) \left({y}^{n}\right)$

$\therefore \left(m n\right) {\left(x + y\right)}^{m n - 1} \left(1 + y '\right) = \left({x}^{m}\right) \left(n {y}^{n - 1} y '\right) + \left(m {x}^{m - 1}\right) \left({y}^{n}\right)$

$\therefore m n {\left(x + y\right)}^{m n - 1} \left(1 + y '\right) = n {x}^{m} {y}^{n - 1} y ' + m {x}^{m - 1} {y}^{n}$

Multiply By $\left(x + y\right) x y$:

$m n {\left(x + y\right)}^{m n} x y \left(1 + y '\right) = \left(x + y\right) \left(n x {x}^{m} {y}^{n} y ' + m {x}^{m} y {y}^{n}\right)$

On the LHS, replace ${\left(x + y\right)}^{m n} = {x}^{m} {y}^{n}$

$m n {x}^{m} {y}^{n} x y \left(1 + y '\right) = \left(x + y\right) {x}^{m} {y}^{n} \left(n x y ' + m y\right)$

Cancel ${x}^{m} {y}^{n}$:

$m n x y \left(1 + y '\right) = \left(x + y\right) \left(n x y ' + m y\right)$

$\therefore m n x y + m n x y y ' = n {x}^{2} y ' + m x y + n x y y ' + m {y}^{2}$

Collect $y '$ terms:

$m n x y y ' - n {x}^{2} y ' - n x y y ' = + m x y + m {y}^{2} - m n x y$

$\therefore n x \left(m y - x - y\right) y ' = m y \left(x + y - n x\right)$

$y ' = \frac{m y \left(x + y - n x\right)}{n x \left(m y - x - y\right)}$