We started with two examples to practice the power rule for derivatives.

Let \(p(x) = \frac{1}{3}x^3 - x^2 - 8x + 24\). Find \(p'(x)\) and find the points where \(p'(x) = 0\).

Let \(y = \dfrac{1}{2x^2} - \dfrac{6}{x}\). Find \(y'\) and the point where \(y' = 0\).

Although derivatives work term-by-term, they don’t play nice with factors. For example, you can’t just take the derivatives the two factors in the expression \(x^2 \cdot x^3\). To work with factors, you need to use the product and quotient rules:

**Product Rule**\(\displaystyle \frac{d}{dx} f(x) g(x) = f'(x) g(x) + f(x) g'(x)\).**Quotient Rule**\(\displaystyle \frac{d}{dx} \frac{f(x)}{g(x)} = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2} = \frac{\text{Lo DHi}- \text{Hi DLo}}{\text{Lo Lo}}\).

We did the following examples:

\(\displaystyle \frac{d}{dx} (3x^3 - 2x + 5) (\sqrt{x} + 2x )\)

\(\displaystyle \frac{d}{dx} (2x+1) (3x-2)\)

\(\displaystyle \frac{d}{dt} \frac{2t}{3t^2 + 16}\)

\(\displaystyle \frac{d}{dx} \frac{1}{x^2 + 4x + 4}\)

\(\displaystyle \frac{d}{dx} \frac{2x-3}{x^3}\)

Today we looked at examples of derivatives in word problems. We did the following examples in class.

- A large tank contains 8 kg of a chemical disolved in 50 L of water. A tap is filling the tank at 5 L/min.
Find a formula for the amount of water in the tank \(W(t)\) as a function of time \(t\).

Find a formula for the concentration \(C(t)\) (recall the concentration of a chemical disolved in water is the mass of the chemical divided by the volume of water.

Calculate \(C'(t)\).

Find \(C'(4)\) and explain what it means (including units).

- Oil producing countries like Russian and Saudi Arabia can affect the price of oil by pumping more (or less oil). Suppose the price of oil is \(\displaystyle P(x) = \frac{200}{\left(1+\frac{1}{10}x\right)}\) where \(x\) measures the daily amount of oil one country produces (in millions of barrels).
Find \(P'(x)\)

If this country is currently producing 10 million barrels of oil each day, then approximately how much will the price rise if they increase production by 1 million additional barrels?

- Continuing the last problem, revenue is price times quantity. Find the revenue function \(R(x)\) for this country as a function of the quantity of oil they produce, \(x\). Then,
Find \(R'(x)\).

Based on the derivative, would this country’s oil revenue increase if they cut production?

The area of a circle is \(A = \pi r^2\). What is \(\dfrac{dA}{dr}\)? Notice that the answer is the circumference of a circle. We talked about why this makes sense in class.

For a company producing a quantity \(x\) of goods to sell, the total cost is a function \(TC(x)\). The average cost is \(AC = \dfrac{TC(x)}{x}\), and the marginal cost is \(MC = TC'(x)\). We used the quotient rule to show that the minimum of the average cost function always occurs when \(MC = AC\).

Today we introduced the **chain rule**. This is the last major derivative rule and it is one of the most important. It takes some practice to get the hang of the chain rule. Here are two video examples:

We did the following examples in class:

\(\displaystyle \frac{d}{dx} ( 4x^3 + 15x)^{20}\).

Let \(y=\sqrt{9-x}\). What is the slope of the tangent line at the point \((5,2)\)?

Find \(f'(x)\) when \(f(x) = \sqrt{25 - x^2 }\).

\(\dfrac{d}{dx} \dfrac{1}{(4-x)^5}\).

We also talked about a second version of the chain rule. If we have a function \(f(g(x))\), then

\[\frac{df}{dx} = \frac{df}{dg} \, \frac{dg}{dx}.\]

A rock is dropped in a still pond and ripples spread outward at a rate of 5 feet per second. How fast is the area of disturbed water increasing after 4 seconds?

An environmental study estimates that for small towns, carbon monoxide levels in the air (in parts per million) are given by \[C(p) = \sqrt{0.5p^2 + 17},\] where \(p\) is the population of the town in thousands. Suppose one town is growing, and it is estimated that the population \(t\) years from now will be \[p(t) = 3.1 + 0.1t^2.\] Find and interpret \(\dfrac{dC}{dt}\) four years from now.