Question #47376

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Answer 1 · 2015-11-06T18:21:27

$y'=-\frac{5x+3y}{3x+5y}$

$5x^2+5y^2+6xy=16$

$d/dx(5x^2+5y^2+6xy)=d/dx(16)$

The derivative of the right part is $0$ because it's a constant. The derivative of the left part is:

$10x+10yy'+6y+6xy'$

So we have:
$10x+10yy'+6y+6xy'=0$

Let's reduce the equation:
$5x+5yy'+3y+3xy'=0$

Now let's factor $y'$
$5x+y'(5y+3x)+3y=0$

Solving for $y'$ :
$y'(5y+3x)=-5x-3$

$y'=\frac{-5x-3y}{5y+3x}$

$y'=-\frac{5x+3y}{3x+5y}$