Calculus II Lesson 3: Logarithmic Functions and Inverse Trig Functions

  1. Implicit Differentiation
  2. Logarithms
  3. Inverse Trig
  4. Homework

Implicit Differentiation

During this lesson, you will learn integrals that result in logarithms and inverse trig functions. These functions are all inverse functions, and so we should first review how to find the derivatives of inverse functions. We do so using a technique, from Calc 1, known as implicit differentiation.

You may wish to first review the topic from the Calculus 1 textbook: Section 3.8: Implicit Differentiation.

Logarithms

A logarithm is the inverse of an exponential function. That is, if $y = b^x$, then $x = \log_b(y)$, and vice versa: if we are trying to graph $y = \log_b(x)$ for some base $b$, we really should think of this as looking for $(x, y)$ values where $b^y = x$.

That is: $\log_b(x)$ and $b^x$ are inverses. If a point $(x, y)$ is on the graph of one of these functions, then switching $x$ and $y$, we get a point on the other graph. For example, if $y = 2^x$, we know that $(3, 8)$ is on the graph of this function. That means $(8, 3)$ is on the graph of $y = \log_2(x)$. That is: $\log_2(8) = 3$.

The natural logarithm function is $f(x) = \log_e(x)$. It is used so often that it has a special name: $\ln(x)$ means $\log_e(x)$. Again, the same general rule applies: $y = \ln(x)$ means $e^y = x$.

To find the derivative of $y = \ln(x)$, it is easier to reason implicitly using the equation $e^y = x$. Take the derivative of both sides with respect to $x$, and we get $e^y y^\prime = 1$. Then solve for $y^\prime$ and get $y^\prime = \dfrac{1}{e^y}$. Remember, though, that we used the equation $e^y = x$, so that means we can substitute $x$ in for $e^y$: that is, $y^\prime = \dfrac{1}{x}$!

This is covered in the excellent 3Blue1Brown video on Implicit Differentiation. Take a look at that one here, as it may help refresh your memory on this topic.

Exercise: Knowing that $\tan(x) = \frac{\sin(x)}{\cos(x)}$, can you find an antiderivative $\int \tan(x) dx$? Hint: use substitution. The natural log will show up here!

Inverse Trig

$y = \arctan(x)$ means $\tan(y) = x$ and $-\pi/2 < y < \pi/2$. (This is literally the definition of $\arctan(x)$: it’s the inverse of $y = \tan(x)$, which means we switch x and y, but we need to restrict the range since $\tan(x)$ is periodic.)

Taking the derivative, implicitly, we get that $1 = (\sec(y))^2 y^\prime$. Since $\sec(y)$ means $\dfrac{1}{\cos(y)}$, we can multiply both sides by $(\cos(y))^2$ and get $y^\prime = (\cos(y))^2$. This is where things get tricky. How do we plug back in $y = \arctan(x)$ into $(\cos(y))^2$ and simplify? Let’s look at a triangle:

x = tan(theta) picture

Let’s think about the relationship between the trig functions, inverse trig functions, and angles:

So if $y = \arctan(x)$, $y$ is the angle that we need to get $\tan(y) = x$ as our output. From the picture, if we want $\tan(\theta) = x$, we can draw a right triangle, with an angle $\theta$, whose opposite side is $x$ and adjacent side is $1$. Then the hypotenuse of the triangle (using the Pythagorean Theorem) is $\sqrt{x^2 + 1}$. In this case, we are looking for cosine of that angle, which would be $\dfrac{1}{\sqrt{x^2+1}}$. So:

\[y^\prime = (\cos(y))^2 = \left(\frac{1}{\sqrt{x^2+1}}\right)^2 = \frac{1}{x^2 + 1}\]

Now we have an explicit derivative of $\arctan(x)$!

On your own: read through Section 1.7 of our textbook, to see the rules for integrals resulting in arctan, arcisn, and arcsec.

Homework

Due Monday, February 6 in class: