Verified Answer :

LammettHash

\(x^2+xy=8\)

Differentiating both sides with respect to \(x\) yields

\(2x+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{2x+y}x\)

Now,

\(\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}\)

So you have

\(-\dfrac{2x+y}x=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}\)

When \(x=2\) and \(y=2\), you have \(\dfrac{\mathrm dx}{\mathrm dt}=-5\), which means

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

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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