We started today with some examples of the chain rule.

\(\displaystyle \frac{d}{dx} (3x^2 - 2x + 1)^4\)

\(y = x \sqrt{1-3x}\). Find \(y'\).

A coffee importer estimates they will sell \(\displaystyle D(p) = \frac{4000}{p^2}\) pounds of coffee if the price is \(p\) dollars. The price of coffee over the next \(t\) weeks is expected to be \(p(t) = 0.02t^2+0.1t+6\).

Find \(\dfrac{dp}{dt}\) when \(t=10\), including the units.

Find \(p(t)\) when \(t=10\).

Find \(\dfrac{dD}{dp}\) when \(t=10\).

Use the chain rule to find \(\dfrac{dD}{dt} = \dfrac{dD}{dp} \, \dfrac{dp}{dt}\).

Next we talked about the concept of concavity. A function \(y = f(x)\) is concave up if \(y'' > 0\) and it is concave down when \(y'' < 0\). We used the 2nd derivative and the graphs of the following functions to see where they are concave up or concave down. We also found the inflection points which are places where the concavity changes from up to down or vice versa.

\(f(x) = \dfrac{1}{x}\)

\(y = \sqrt{x}\)

\(y = x^3 - 6x^2 + x - 5\)

\(f(x) = 2x^6 - 5x^4 + 7 x - 3\)

The function \(f(x) = \dfrac{1}{1+x^2}\) has a graph that is shaped like a bell. It has two inflection points. Use the second derivative \(f''(x) = \dfrac{2(3x^2-1)}{(x^2+1)^3}\) to find the inflection points.

Wednesday, March 24

Today we started with a quick review self-quiz for midterm 2. It included the following derivatives problems:

Find \(\dfrac{d}{dx} \, \dfrac{4}{x^2}\).

Find \(\dfrac{d}{dx} \, x \sqrt{x}\).

Find \(\dfrac{d}{dx} \, \dfrac{x^2}{x+1}\).

Find \(\dfrac{d}{dx} \, \sqrt{x^4+x^2}\).

Find the equation for the tangent line to the parabola \(y = 3x-x^2\) at \((2,2)\).

Find the inflection point of \(y = x^3 - 6x^2 - 16x\).

We also did a few limit examples and some of the examples from the review problems.

Friday, March 26

Today we used derivatives to find intervals of increase & decrease and also local maximum & minimums. We did the following examples.

\(f(x) = x^3 - 6x^2 + 9x + 2\)

\(y = x^4 + 8x^3 + 18x^2 - 8\)

\(\displaystyle f(x) = \frac{x^2}{x-2}\)

\(\sqrt{3 - 2t - t^2}\)

\(y = (4x-x^2)^{2/3}\)

Suppose that a business has total costs of \(C(x) = 2x^2 + 3x + 5\) and revenue \(R(x) = 5x-2x^2\). What level of production \(x\) would maximize profit (recall that profit is revenue minus costs)?