Respuesta :

[tex]x^2+xy=8[/tex]

Differentiating both sides with respect to [tex]x[/tex] yields

[tex]2x+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{2x+y}x[/tex]

Now,

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]

So you have

[tex]-\dfrac{2x+y}x=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]

When [tex]x=2[/tex] and [tex]y=2[/tex], you have [tex]\dfrac{\mathrm dx}{\mathrm dt}=-5[/tex], which means

[tex]-\dfrac{2(2)+2}2=-\dfrac{\frac{\mathrm dy}{\mathrm dt}}5[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dt}=15[/tex]

The differentiation(dy/dt) of y with respect to t of the given function with all the given parameters gives us;

dy/dt = 15

We are given the function;

x² + xy = 8

Differentiating with respect to x gives;

2x + x(dy/dx) + y = 0

Making dy/dx the subject gives;

dy/dx = -(y + 2x)/x

Now, we are told that x = x(t) and y = y(t) are both functions of t. This means that;

dy/dx = (dy/dt)/(dx/dt)

  • We are given dx/dt = -5, thus plugging in relevant values, we have;

-(y + 2x)/x = (dy/dt)/-5

At x = 2 and y = 2, we have;

-(2 + 2(2))/2 = (dy/dt)/-5

-3 = (dy/dt)/-5

dy/dt = -3 * -5

dy/dt = 15

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