Calculus Word Problem: Several Variables Demand for a certain kind of SUV obeys the following equation,

$$D(x, y) = 21000 - \frac {\sqrt{x}}{2} - 11(0.3y-10)^{3/2}$$

where x is the price per car in dollars, y is the cost of gasoline per litre, and D is the number of cars.

Suppose that the price of the car and the price of gasoline t years from now obey the following equations:

$$x  =  55,200 + 100t ~~~~and~~~~      y = 136 + 10 \sqrt{t}$$  

What will the rate of change of the demand be (with respect to time) 5 years from now?

I would really appreciate getting some help answering this question. I truly don't know where to start with this question never mind the solution. What practice or formula should I be using to answer this question?

New info

First I plug t= 5 into both equations. so

$x= 55700$ 

$y=158$

But is still don't know what to do after...
You need to differentiate the demand equation.  It is a bit more challenging here because of the fractional exponents.  You find the derivative either by using partial derivatives, as suggested by @callculus, or you can implicitly differentiate with respect to time.  I find that the latter approach gets a little confusing here, so, in this instance, I prefer the former.  The former approach allows you to break it into "smaller pieces", though the end result is the same.

You are also explicitly given information on how $x$ and $y$ are functions of time.  You can calculate the value of $x$ and $y$ at time = 5 using those functions, and you can calculate the values of $dx \over dt$ and $dy \over dt$ by taking the derivative of those functions.

Implicit Differentiation with respect to time

To implicitly differentiate the demand equation with respect to time, you first differentiate each variable with respect to itself, and then multiply by the derivative of that variable with respect to time.  This is the chain rule.

Because the the demand equation consists of the sum of two smaller expressions, the derivative sum rule says that we can simply add the derivatives of each expression.  That is,

$$\frac {d (u + v)}{dx} = \frac {du}{dx} + \frac {dv}{dx}$$

So, let's first differentiate $21000 - \frac {\sqrt{x}}{2}$ with respect to $x$.  You can rewrite that as $21000 - \frac {1}{2}x^{1/2}$.  Differentiating that gives you $- \frac {1}{4 \sqrt{x}}$.  We are differentiating with respect to time, therefore we need to multiply by the derivative of the inner function, $\frac {dx}{dt}$.

Now let's differentiate the second expression, $-11(0.3y-10)^{3/2}$, with respect to $y$.  If I have not made an error, the derivative is $- \frac {99}{20}*\sqrt{0.3y-10}$.  Again, we are differentiating with respect to time, therefore we need to multiply by the derivative of the inner function, $\frac {dy}{dt}$.  

Put it all together for the derivative of the demand equation:

$$D(x,y) = 21000 - \frac {\sqrt{x}}{2} - 11(0.3y-10)^{3/2}$$
$$\frac {dD}{dt} = (- \frac {1}{4 \sqrt{x}} \frac {dx}{dt}) - (\frac {99}{20}*\sqrt{0.3y-10}* \frac {dy}{dt})$$

We have one equation and 5 unknowns, $\frac {dD}{dt}, x, y, \frac {dx}{dt} ~and~ \frac {dy}{dt} $.  We are attempting to solve for $\frac {dD}{dt}$, so we just need to figure out values for $x, y, \frac {dx}{dt} ~and~ \frac {dy}{dt}$.  As noted above, we have the formulas for $x$ and $y$. Therefore we can solve for the values of $x$ and $y$ at $t=5$, and we can differentiate those equations to find $\frac {dx}{dt}$ and $\frac {dy}{dt}$.

The values of $x$ and $y$ at $t=5$ are

$$x = 55200 + 100t$$
$$x = 55200 + 100(5)$$
$$x = 55700$$

$$y = 136+10 \sqrt{t}$$
$$y = 136+10 \sqrt{5}$$
$$y \approx 158.36$$

The derivatives of $x$ and $y$ with respect to time are:

$$\frac {dx}{dt} = 100$$
$$\frac {dy}{dt} = \frac {5}{\sqrt{t}}$$

Substitute known values in to the equation:
$$\frac {dD}{dt} = - \frac {1}{4 \sqrt{x}} *100 - \frac {99}{20}*\sqrt{0.3y-10}*\frac {dy}{dt}$$

$$\frac {dD}{dt} = - \frac {1}{4 \sqrt{55700}} *100 - \frac {99}{20}*\sqrt{0.3(158.36)-10}*\frac {5}{\sqrt{5}}$$

If I've made no errors, the result is

$$\frac {dD}{dt} \approx -67.89$$

Partial Derivatives

When taking partial derivatives, you explicitly break down the process into smaller steps, but it is exactly the procedure outlined above.  You take the derivative with respect to the variable itself (for example $x$) and you treat the other variable (in this case $y$) as constant.  So, when calculating the partial derivative with respect to $x$ the equation effectively becomes
$$D(x,y) = 21000 - \frac {\sqrt{x}}{2} - C$$
$$\frac {\partial D}{\partial x} = (- \frac {1}{4 \sqrt{x}}) - 0$$

And, similarly, when calculating the partial derivative with respect to $y$, the equation effectively becomes
$$D(x,y) = C - 11(0.3y-10)^{3/2}$$
$$\frac {\partial D}{\partial y} = 0 - (\frac {99}{20}*\sqrt{0.3y-10})$$

Note that the slightly different notation $\partial$ is used to signify a partial derivative.

Once the partials are calculated, you use the chain rule formula as suggested by @callculus to calculate the derivative of demand with respect to time, that is:

$$ \frac {dD}{dt} = \frac {\partial D}{\partial x} \frac {dx}{dt} + \frac {\partial D}{\partial y} \frac {dy}{dt}$$

which, of course, is exactly the same result that you would obtain with implicit differentiation.
Hint: The change of $D$ at $t$ is

$$\frac{dD}{dt}=\frac{\partial D}{\partial x}\cdot \frac{d x}{dt}+\frac{\partial D}{\partial y}\cdot \frac{dy}{dt}$$

In case of $\frac{dD}{dx}$ you differentiate $D(x,y)$ w.r.t $x$. After that you calculate the value of $x(t)$ at $t=5$. And then you insert that value into $\frac{dD}{dx}$.