In-Depth Notes on Derivatives of Inverse Trigonometric and Logarithmic Functions

Let’s explore the derivatives of inverse trigonometric functions together! Don’t worry if some of this seems unfamiliar; we'll go step-by-step through each part.

What Are Inverse Trigonometric Functions?

First, let's understand what inverse trigonometric functions are. They are functions that give us the angle corresponding to a particular sine, cosine, or tangent value. For example:

  • If you know that
    \sin(\theta) = x, then \arcsin(x) = \theta.

Key Derivatives

Now, let's focus on how we can differentiate these functions. Here are some key derivatives you need to remember:

  1. arcsin(x)

    • Derivative: When you want to differentiate arcsin(x), you use the following formula:
      \frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1 - x^2}}

    • Domain: This means we only apply this when -1 < x < 1 to ensure we're working with valid inputs.

  2. arccos(x)

    • Derivative: Similarly, for arccos(x), the derivative is:
      \frac{d}{dx} \arccos x = -\frac{1}{\sqrt{1 - x^2}}

    • Domain: Again, this is valid for -1 < x < 1.

  3. arctan(x)

    • Derivative: For the arctangent function, the derivative is a bit simpler:
      \frac{d}{dx} \arctan x = \frac{1}{1 + x^2}

Examples of Differentiation

Let's practice by going through some examples:

Example 1

Find \frac{dy}{dx} if y = \arcsin x.

  • Use the derivative formula we learned:
    \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}.

Example 2

Find \frac{dy}{dx} if y = \arctan x.

  • Here, we'll apply the arctan derivative:
    \frac{dy}{dx} = \frac{1}{1 + x^2}.

Example 3

Let’s differentiate a function:
f(x) = (\arctan x)^2.

  • To do this, use the chain rule:

    • Let u = \arctan x. So,

    • f(x) = u^2.

    • Differentiate using the chain rule:
      \frac{df}{dx} = 2u \cdot \frac{du}{dx} = 2\arctan x \cdot \frac{1}{1 + x^2}.

Example 4

Differentiate the function:
g(x) = \sqrt{1 - x^2} \arccos x.

  • Use the product rule since this is a product of two functions. Let:

    • u = \sqrt{1 - x^2}

    • v = \arccos x

  • So,
    \frac{dg}{dx} = u'v + uv' (where u' is the derivative of u and v' is the derivative of v).

Remember!

It's essential to not only memorize these formulas but to also practice applying them in different contexts. With time, you'll become more comfortable and quicker with these derivatives! Keep practicing, and don't hesitate to ask questions if you're unsure about something!