Calculus I Lesson 9: Trig Derivatives + Quotient Rule

  1. Warm Up
  2. Derivatives of sine and cosine
    1. Higher derivatives of sin and cosine
  3. Quotient Rule
  4. Reminders

Warm Up

Consider the graph of $y = \sin(x)$.

  1. At what points is $y^\prime = 0$?
  2. Where is $y^\prime > 0$?
  3. Where is $y^\prime < 0$?

Using this information, sketch a rough graph of $y^\prime$. Does this look like any function you know of already?

Derivatives of sine and cosine

First watch this video, where I go through the derivatives of $\sin(x)$ and $\cos(x)$:

Summary:

Proof: We will only look at the derivative of $\sin(x)$ here, but the idea is the same for the derivative of $\cos(x)$. Use the definition of the derivative: We look at the limit as $h \rightarrow 0$ of $\dfrac{\sin(x + h) - \sin(x)}{h}$. Use angle sum formula so that $\sin(x + h) = \sin(x)\cos(h) + \cos(x) \sin(h)$. Then we get:

\[\frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}\]

Combine the expressions that have $\sin(x)$ in them:

\[\frac{\sin(x)(\cos(h) - 1)}{h} + \frac{\cos(x)\sin(h)}{h}\]

As $h \rightarrow 0$, we’ve seen that $\frac{\sin(h)}{h} \rightarrow 1$. We did not see this, but if you look at the graph of $\frac{\cos(x) - 1}{x}$, as $x \rightarrow 0$, that approaches 0. So as $h \rightarrow 0$, this whole expression approaches $\cos(x)$!

Exercises: Find the derivatives of the following functions, using the product rule and the trig derivatives above:

  1. $y = x^2 \sin(x)$
  2. $y = \cos(x)\sin(x)$
  3. $y = (\cos(x))^2$. Recall that this is $\cos(x)\cos(x)$.
Check your answers:
  1. $2x\sin(x) + x^2 \cos(x)$
  2. $-(\sin(x))^2 + (\cos(x))^2$, or $(\cos(x))^2 - (\sin(x))^2$
  3. $-2\sin(x)\cos(x)$

Higher derivatives of sin and cosine

Notice the following: let $y = \sin(x)$

etc.

Exercises:

  1. What is the 100th derivative of $\sin(x)$?
  2. What is the 101st derivative of $\sin(x)$?
  3. Explain the pattern of the derivatives of sin and cosine as best as you can.
  4. Have you seen any other patterns like this before?

Now take a look at the following video:

Quotient Rule

Watch this video where I go over the quotient rule:

To recap: if we have two functions $f(x)$ and $g(x)$, and we know that they are both differentiable (at some point $x = a$)j, we can figure out the derivative of $h(x) = \dfrac{f(x)}{g(x)}$, assuming that $g(x) \neq 0$ at that point. The quotient rule is:

\[(\dfrac{f(x)}{g(x)})^\prime = \dfrac{f^\prime(x) g(x) - g^\prime(x) f(x)}{(g(x))^2}\]

As an aside: the method we used to figure this out is called implicit differentiation. We will be revisiting this method often: if we have an equation with $x$ and $y$ in it, and we don’t quite know the derivative of $y$ by itself, we sometimes can differentiate implicitly. This means that we leave a placeholder, $y^\prime$, for the derivative.

In this example, if $y = \dfrac{f(x)}{g(x)}$, we don’t know $y^\prime$. But, if we multiply both sides by $g(x)$, we know that $(y\cdot g(x))^\prime = f^\prime(x)$. Then we can use the product rule on the left hand side:

\[y^\prime g(x) + g^\prime(x) y = f^\prime(x)\]

We want to get $y^\prime$ by itself. So subtract $g^\prime(x) y$:

\[y^\prime g(x) = f^\prime(x) - g^\prime(x)y\]

Now divide both sides by $g(x)$:

\[y^\prime = \dfrac{f^\prime(x)}{g(x)} - \dfrac{g^\prime(x) y}{g(x)}\]

Replace $y$ by $\dfrac{f(x)}{g(x)}$:

\[y^\prime = \dfrac{f^\prime(x)}{g(x)} - \dfrac{g^\prime(x) f(x)}{(g(x))^2}\]

And now get a common denominator to finish the problem:

\[y^\prime = \dfrac{f^\prime(x) g(x) - g^\prime(x) f(x)}{(g(x))^2}\]

Exercises:

  1. Find the derivative of $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$.
  2. Find the derivative of $\sec(x) = \dfrac{1}{\cos(x)}$.

Reminders

DeltaMath HW 6 due on Monday.