Given $x^{t} \cdot y^{m} - {(x + y)}^{m + t} = 0$ $x^t * y^m - (x+y)^(m+t)=0$ determine $\frac{d y}{d x}$ $dy/dx$ ?

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Given $x^{t} \cdot y^{m} - {(x + y)}^{m + t} = 0$ $x^t * y^m - (x+y)^(m+t)=0$ determine $\frac{d y}{d x}$ $dy/dx$ ?

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Answer 1 · 2016-10-28T09:00:19

See below.

Explanation:

$f (x, y) = x^{t} \cdot y^{m} - {(x + y)}^{m + t} = 0$ $f(x,y)=x^t * y^m - (x+y)^(m+t)=0$

$d f = f_{x} d x + f_{y} d y = 0$ $df = f_xdx+f_ydy=0$ so

$\frac{d y}{d x} = - \frac{f_{x}}{f_{y}}$ $dy/dx=-f_x/(f_y)$ but

$f_{x} = \frac{t}{x} x^{t} y^{m} - \frac{t + m}{x + y} {(x + y)}^{t + m}$ $f_x=t/x x^ty^m-(t+m)/(x+y)(x+y)^(t+m)$

and

$f_{y} = \frac{m}{y} x^{t} y^{m} - \frac{t + m}{x + y} {(x + y)}^{t + m}$ $f_y=m/y x^ty^m-(t+m)/(x+y)(x+y)^(t+m)$

but

$x^{t} y^{m} = {(x + y)}^{t + m}$ $x^ty^m=(x+y)^(t+m)$ then

$f_{x} = (\frac{t}{x} - \frac{t + m}{x + y}) x^{t} y^{m}$ $f_x=(t/x-(t+m)/(x+y))x^ty^m$ and
$f_{y} = (\frac{m}{y} - \frac{t + m}{x + y}) x^{t} y^{m}$ $f_y=(m/y-(t+m)/(x+y))x^ty^m$

so

$\frac{d y}{d x} = - \frac{\frac{t}{x} - \frac{t + m}{x + y}}{\frac{m}{y} - \frac{t + m}{x + y}} = \frac{y}{x}$ $dy/dx=-(t/x - (t + m)/(x + y))/(m/y - (t + m)/(x + y))=y/x$

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Given $x^{t} \cdot y^{m} - {(x + y)}^{m + t} = 0$ $x^t * y^m - (x+y)^(m+t)=0$ determine $\frac{d y}{d x}$ $dy/dx$ ?

1 Answer

Explanation:

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