4.3 Derivatives of Inverse Trigonometric Functions

4.3 Derivatives of Inverse Trigonometric Functions

Quick Answer: The derivatives of inverse trigonometric functions yield purely algebraic expressions. The three essential formulas for the CLEP exam are: $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}}$, $\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2}$, and $\frac{d}{dx}[\text{arcsec } x] = \frac{1}{|x|\sqrt{x^2 - 1}}$. Each "co-" inverse trigonometric derivative is simply the negative of its co-function counterpart. For composite functions, multiply the numerator by the inner derivative $g'(x)$.

---

1. Master Table of Inverse Trigonometric Derivatives

The six inverse trigonometric functions are defined on restricted domains of their trigonometric parents to ensure they pass the horizontal line test and remain invertible bijective functions.

The Co-Function Negative Symmetry

Notice the perfect structural pairing in the table: $\frac{d}{dx}[\arccos x] = -\frac{d}{dx}[\arcsin x]$ $\frac{d}{dx}[\text{arccot } x] = -\frac{d}{dx}[\arctan x]$ $\frac{d}{dx}[\text{arccsc } x] = -\frac{d}{dx}[\text{arcsec } x]$ This symmetry arises directly from the complementary angle identity $\arcsin x + \arccos x = \frac{\pi}{2}$. Taking the derivative of both sides gives $\frac{d}{dx}[\arcsin x] + \frac{d}{dx}[\arccos x] = 0$, proving that their rates of change are exact opposites.

---

2. Geometric Right-Triangle and Implicit Derivations

Understanding the derivation of these formulas prevents memorization errors and reinforces implicit differentiation skills.

Derivation 1: Proving $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}}$

1. Let $y = \arcsin x$, which means $\sin y = x$ for $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $x \in [-1, 1]$.
2. Differentiate both sides implicitly with respect to $x$: $\frac{d}{dx}[\sin y] = \frac{d}{dx}[x] \implies \cos y \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\cos y}$
3. Construct a reference right triangle with angle $y$:
   • $\text{Opposite side} = x$
   • $\text{Hypotenuse} = 1$
   • By Pythagorean theorem: $\text{Adjacent side} = \sqrt{1^2 - x^2} = \sqrt{1 - x^2}$
4. From the triangle, $\cos y = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \sqrt{1 - x^2}$. (Since $y \in [-\pi/2, \pi/2]$, $\cos y \ge 0$).
5. Substitute back: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \quad \blacksquare$

                     Right Triangle for y = arcsin(x)

                                  /|
                                 / |
                     Hypotenuse /  |  Opposite = x
                         = 1   /   |
                              / y  |
                             /_____|
                          Adjacent = √(1 - x²)

Derivation 2: Proving $\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2}$

1. Let $y = \arctan x \implies \tan y = x$ for $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
2. Differentiate implicitly: $\sec^2 y \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\sec^2 y}$
3. Use the fundamental Pythagorean identity $\sec^2 y = 1 + \tan^2 y$: $\frac{dy}{dx} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2} \quad \blacksquare$

Derivation 3: Proving $\frac{d}{dx}[\text{arcsec } x] = \frac{1}{|x|\sqrt{x^2 - 1}}$

1. Let $y = \text{arcsec } x \implies \sec y = x$ for $y \in [0, \pi/2) \cup (\pi/2, \pi]$.
2. Differentiate implicitly: $\sec y \tan y \cdot y' = 1 \implies y' = \frac{1}{\sec y \tan y}$.
3. Using the identity $\tan^2 y = \sec^2 y - 1 = x^2 - 1$, we have $\tan y = \pm \sqrt{x^2 - 1}$.
4. Because the slope of the secant inverse function is strictly positive across its entire domain, the absolute value $|x|$ ensures the denominator remains positive for all $|x| > 1$: $\frac{d}{dx}[\text{arcsec } x] = \frac{1}{|x|\sqrt{x^2 - 1}} \quad \blacksquare$

---

3. Chain Rule Generalized Forms

When the argument is a differentiable function $u = g(x)$, combine the inverse trigonometric formulas with the Chain Rule by placing $g'(x)$ in the numerator:

$\frac{d}{dx}[\arcsin(g(x))] = \frac{g'(x)}{\sqrt{1 - [g(x)]^2}}$ $\frac{d}{dx}[\arctan(g(x))] = \frac{g'(x)}{1 + [g(x)]^2}$ $\frac{d}{dx}[\text{arcsec}(g(x))] = \frac{g'(x)}{|g(x)|\sqrt{[g(x)]^2 - 1}}$

Step-by-Step Worked Examples

Example 1: Arctangent with Polynomial Argument

Find $f'(x)$ for $f(x) = \arctan(3x^2)$.

• $g(x) = 3x^2 \implies g'(x) = 6x$
• $[g(x)]^2 = (3x^2)^2 = 9x^4$
• $f'(x) = \frac{6x}{1 + 9x^4}$

Example 2: Arcsine with Exponential Argument

Find $y'$ for $y = \arcsin(e^{2x})$.

• $g(x) = e^{2x} \implies g'(x) = 2e^{2x}$
• $[g(x)]^2 = (e^{2x})^2 = e^{4x}$
• $y' = \frac{2e^{2x}}{\sqrt{1 - e^{4x}}}$

Example 3: Arctangent with Radical Argument

Find $\frac{dy}{dx}$ for $y = \arctan(\sqrt{x})$.

• $g(x) = x^{1/2} \implies g'(x) = \frac{1}{2\sqrt{x}}$
• $[g(x)]^2 = (\sqrt{x})^2 = x$
• $\frac{dy}{dx} = \frac{\frac{1}{2\sqrt{x}}}{1 + x} = \frac{1}{2\sqrt{x}(1 + x)}$

Example 4: Product Rule Combined with Inverse Trig

Find the slope of the tangent line to $y = x^2 \arctan(2x)$ at $x = 1/2$.

1. Apply Product Rule: $y' = 2x \arctan(2x) + x^2 \cdot \frac{2}{1 + (2x)^2} = 2x \arctan(2x) + \frac{2x^2}{1 + 4x^2}$.
2. Substitute $x = 1/2$: $y'\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)\arctan(1) + \frac{2(1/4)}{1 + 4(1/4)} = 1 \cdot \left(\frac{\pi}{4}\right) + \frac{1/2}{2} = \frac{\pi}{4} + \frac{1}{4} = \frac{\pi + 1}{4}$

---

4. Algebraic Simplifications of Composite Expressions

A classic CLEP problem type asks to differentiate an expression where an inverse trigonometric derivative collapses into a remarkably simple algebraic result.

Example: Differentiating $f(x) = x \arcsin x + \sqrt{1 - x^2}$

1. Differentiate the first term $x \arcsin x$ using the Product Rule: $\frac{d}{dx}[x \arcsin x] = (1)\arcsin x + x\left(\frac{1}{\sqrt{1 - x^2}}\right) = \arcsin x + \frac{x}{\sqrt{1 - x^2}}$
2. Differentiate the second term $\sqrt{1 - x^2} = (1 - x^2)^{1/2}$ using the Chain Rule: $\frac{d}{dx}\left[(1 - x^2)^{1/2}\right] = \frac{1}{2}(1 - x^2)^{-1/2}(-2x) = -\frac{x}{\sqrt{1 - x^2}}$
3. Sum the derivatives: $f'(x) = \left( \arcsin x + \frac{x}{\sqrt{1 - x^2}} \right) + \left( -\frac{x}{\sqrt{1 - x^2}} \right) = \arcsin x$

---

5. CLEP Traps and Exam Strategies

• Trap 1: Notation Confusion: Inverse Trig vs. Reciprocal Trig. $\arcsin x = \sin^{-1} x \neq (\sin x)^{-1} = \csc x$. The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1-x^2}}$, whereas the derivative of $(\sin x)^{-1}$ is $-\csc x \cot x$. Never confuse inverse functions with multiplicative reciprocals.
• Trap 2: Forgetting to Square the Inner Function. In $\frac{d}{dx}[\arctan(4x)]$, the denominator is $1 + (4x)^2 = 1 + 16x^2$, NOT $1 + 4x^2$.
• Trap 3: Sign Errors on Co-functions. Remember that all inverse functions starting with "arc-co" ($\arccos$, $\text{arccot}$, $\text{arccsc}$) carry a negative sign in their derivatives.
• Trap 4: Boundary Points of Non-Differentiability. At $x = \pm 1$, the function $y = \arcsin x$ is continuous, but the denominator of its derivative $\sqrt{1 - x^2}$ becomes zero. The graph has vertical tangent lines at $(\pm 1, \pm \pi/2)$, meaning the derivative does not exist at the domain boundaries.