The problem ultimately comes down to finding \(\displaystyle \dfrac{dw}{dt}\) which reads as the derivative of w (the hypotenuse of the triangle) in regards to time.

This is absolutely correct.

It is frequently helpful in math to

**start** by figuring out where you want to

**end up** and then figuring out in

**reverse** how to get there. Here are asked to find the rate of change with respect to time of the distance between a fixed location and a moving particle when that particle is at a specific location. That is where you want to end up. Translate that endpoint into the language of calculus. For that we need names for our variables. I arbitrarily chose w for distance and t for time. And the rate of change is

\(\displaystyle \dfrac{dw}{dt}.\)

Now if you just had a function relating t and w, you would differentiate and virtually be done. But there does not seem to be such a function directly. What functional relationships can you see? Well, there must be a relationship between the distance between the location of the fixed point and the location of the moving particle, If we could find that out and the relationship between the location of the moving point and time, we could differentiate both and use the chain rule. But wait, we already know the rate of change of the moving point with respect to time.

\(\displaystyle \dfrac{dx}{dt}\) reads as the derivative of coordinate x in regards to time which is given information. Thus, 2 units/sec.

Correct. Very good. So now we have our plan: figure out the function relating the distance between the locations of the fixed point and the moving particle. Differentiate that, apply the chain rule to get the derivative we want, and plug in the appropriate values.

Function 1 is the equation we created for length of w. Function 2 is time.

The first sentence is correct. The second is not. The second function is not time, but the velocity of the moving particle as evidenced by its units: units of distance per second. The second function is

**already** a derivative. (All derivatives are functions.)

So the final problem before plugging x = 5 is now

(1/2) x^{2}+81^{(-1/2)} times 2x +2

where I differentiated function 1 which needed the chain rule and then added the derivative of time which was already given in the problem.

Now you have jumped the rails.

Yes, you use the the chain rule to differentiate function 1.

\(\displaystyle w = \sqrt{x^2 + 81} \text { and } u = x^2 + 81 \implies w = \sqrt{u} = u^{1/2} \text { and } \dfrac{du}{dx} = 2x \implies\)

\(\displaystyle \dfrac{dw}{du} = \dfrac{1}{2} * u^{(1/2)-1} = \dfrac{1}{2\sqrt{u}} = \dfrac{1}{2\sqrt{x^2 + 81}} \text { and }\)

\(\displaystyle \dfrac{dw}{dx} = \dfrac{dw}{du} * \dfrac{du}{dx} = \dfrac{1}{\cancel 2 \sqrt{x^2 + 81}} * \cancel 2x = \dfrac{x}{\sqrt{x^2 + 81}}.\)

So, except for the cancellations, you did that correctly.

**BUT**
\(\displaystyle \dfrac{dw}{dx} + \dfrac{dx}{dt} \ne \dfrac{dw}{dt}.\)

You do not get meaningful derivatives by adding derivatives that are not with respect to the same thing.

You must apply the chain rule again. (Although the Leibniz notation is only an analogy under standard analysis, it is frequently a helpful analogy.)

\(\displaystyle \dfrac{dw}{dt} = \dfrac{dw}{dx} * \dfrac{dx}{dt} = \dfrac{x}{\sqrt{x^2 + 81}} * 2 = \dfrac{2x}{\sqrt{x^2 + 81}}.\)

Now you can plug in x = 5.