Inverse Trig Derivatives- Worksheets and Practice

Inverse Trig Derivatives: The Complete Practice Guide

Inverse trigonometric derivatives show up in calculus exams, problem sets, and real engineering work. You need to know them. Not "kind of know them." Actually know them.

This guide cuts through the confusion. Six functions. Six formulas. Examples and practice problems with solutions. No fluff.

What Are Inverse Trig Derivatives?

When you take the derivative of an inverse trig function, you're finding the rate of change of an angle with respect to its trig ratio.

For example, if y = arcsin(x), then dy/dx tells you how fast the angle is changing as the sine value changes. The formulas are straightforward—mostly involving the square root of 1 minus something squared.

The Six Inverse Trig Functions and Their Derivatives

Here's the complete list. Memorize this table.

The domain restrictions matter. If you plug in values outside the domain, you're asking for trouble. The derivative formulas assume you're working within valid inputs.

How to Derive These Formulas

You don't need to derive these from scratch every time, but understanding the proof helps them stick.

The arcsin Derivative Proof

Start with y = arcsin(x). Rewrite as sin(y) = x.

Take derivatives of both sides:

cos(y) · dy/dx = 1

Solve for dy/dx:

dy/dx = 1 / cos(y)

Use the identity cos²(y) = 1 - sin²(y), which means cos(y) = √(1 - sin²(y)). Since sin(y) = x:

cos(y) = √(1 - x²)

Therefore:

d/dx arcsin(x) = 1 / √(1 - x²)

The arccos proof follows the same steps. The only difference is that cos(y) = x becomes x = cos(y), and the chain rule flips the sign somewhere in the process. That's why arccos has a negative sign in front.

The arctan Derivative Proof

Start with y = arctan(x), so tan(y) = x.

Derivative of tan(y) is sec²(y) · dy/dx. Set equal to 1:

sec²(y) · dy/dx = 1

dy/dx = 1 / sec²(y)

Since sec²(y) = 1 + tan²(y) and tan(y) = x:

dy/dx = 1 / (1 + x²)

Done. Same logic applies to arccot—you get the negative version.

How to Memorize These Formulas

Here's the pattern that makes memorization easier:

• arcsin and arccos share the denominator √(1 - x²). The difference is the sign.
• arctan and arccot share the denominator (1 + x²). The difference is the sign.
• arcsec and arccsc share the denominator |x|√(x² - 1). The difference is the sign.

The co-functions (cos, cot, csc) always have a negative sign. The principal functions (sin, tan, sec) have a positive sign. That's the pattern. Remember that and you can reconstruct any of them.

One more thing: the derivative of arcsin and arctan look almost identical except for what's inside the radical. arcsin has "1 minus x squared" under a square root. arctan just has "1 plus x squared" in the denominator. Don't mix them up.

Practice Problems and Worksheets

Work through these. Check your answers. If you get stuck, the solutions are below.

Basic Derivatives (Find dy/dx)

1. y = arcsin(3x)
2. y = arctan(2x + 1)
3. y = arccos(x²)
4. y = 5arctan(x) + 3arcsin(x)
5. y = arcsec(x/2)

Chain Rule Problems

1. y = sin(arccos(x))
2. y = arctan(√x)
3. y = arcsin(x) / arctan(x)
4. y = (arccos(x))³
5. y = √(arctan(x))

Harder Composite Functions

1. y = arctan(x² + 1) · arcsin(x)
2. y = ln(arctan(x))
3. y = e^(arctan(x))
4. y = arctan(1/x)
5. y = arcsin(x) + arccos(x)

Solutions and Step-by-Step Work

Basic Derivatives Solutions

Problem 1: y = arcsin(3x)

Use the chain rule. The outer function is arcsin(u) where u = 3x.

dy/dx = (1 / √(1 - (3x)²)) · 3

dy/dx = 3 / √(1 - 9x²)

Problem 2: y = arctan(2x + 1)

Outer function: arctan(u), inner: u = 2x + 1

dy/dx = (1 / (1 + (2x + 1)²)) · 2

dy/dx = 2 / (1 + (2x + 1)²)

Simplify if needed: dy/dx = 2 / (1 + 4x² + 4x + 1) = 2 / (4x² + 4x + 2)

Problem 3: y = arccos(x²)

Remember the negative sign for arccos.

dy/dx = -1 / √(1 - (x²)²) · 2x

dy/dx = -2x / √(1 - x⁴)

Problem 4: y = 5arctan(x) + 3arcsin(x)

Apply each derivative and multiply by coefficients:

dy/dx = 5 · (1 / (1 + x²)) + 3 · (1 / √(1 - x²))

dy/dx = 5 / (1 + x²) + 3 / √(1 - x²)

Problem 5: y = arcsec(x/2)

Let u = x/2. Then arcsec(u) derivative is 1 / (|u|√(u² - 1))

dy/dx = (1 / (|x/2| · √((x/2)² - 1))) · (1/2)

dy/dx = 1 / (|x| · √(x²/4 - 1))

Multiply numerator and denominator by 2:

dy/dx = 2 / (|x| · √(x² - 4))

Chain Rule Problems Solutions

Problem 1: y = sin(arccos(x))

This simplifies first. arccos(x) gives an angle whose cosine is x. sin(arccos(x)) = √(1 - x²).

dy/dx = d/dx √(1 - x²) = (1 / 2√(1 - x²)) · (-2x)

dy/dx = -x / √(1 - x²)

Problem 2: y = arctan(√x)

Let u = √x = x^(1/2). dy/dx = (1 / (1 + (√x)²)) · (1 / 2√x)

dy/dx = (1 / (1 + x)) · (1 / 2√x)

dy/dx = 1 / (2√x(1 + x))

Problem 3: y = arcsin(x) / arctan(x)

Quotient rule. Let u = arcsin(x), v = arctan(x)

u' = 1/√(1-x²), v' = 1/(1+x²)

dy/dx = (v·u' - u·v') / v²

dy/dx = ((arctan(x) / √(1-x²)) - (arcsin(x) / (1+x²))) / (arctan(x))²

Problem 4: y = (arccos(x))³

Chain rule twice. Outer: u³, middle: arccos(u), inner: u = x

dy/dx = 3(arccos(x))² · (-1/√(1-x²))

dy/dx = -3(arccos(x))² / √(1-x²)

Problem 5: y = √(arctan(x))

Rewrite as (arctan(x))^(1/2)

dy/dx = (1/2)(arctan(x))^(-1/2) · (1/(1+x²))

dy/dx = 1 / (2√(arctan(x)) · (1+x²))

Harder Composite Functions Solutions

Problem 1: y = arctan(x² + 1) · arcsin(x)

Product rule. u = arctan(x²+1), v = arcsin(x)

u' = 2x / (1 + (x²+1)²) = 2x / (1 + x⁴ + 2x² + 1) = 2x / (x⁴ + 2x² + 2)

v' = 1/√(1-x²)

dy/dx = u'v + uv'

dy/dx = (2x / (x⁴ + 2x² + 2)) · arcsin(x) + arctan(x²+1) / √(1-x²)

Problem 2: y = ln(arctan(x))

Chain rule with logarithm.

dy/dx = (1 / arctan(x)) · (1 / (1+x²))

dy/dx = 1 / (arctan(x) · (1+x²))

Problem 3: y = e^(arctan(x))

Chain rule with exponential.

dy/dx = e^(arctan(x)) · (1 / (1+x²))

Problem 4: y = arctan(1/x)

Let u = 1/x = x^(-1)

dy/dx = (1 / (1 + (1/x)²)) · (-1/x²)

dy/dx = (1 / (1 + 1/x²)) · (-1/x²)

Simplify the denominator: 1 + 1/x² = (x² + 1) / x²

dy/dx = (x² / (x² + 1)) · (-1/x²)

dy/dx = -1 / (x² + 1)

Notice this is similar to the derivative of arccot(x). That's not a coincidence.

Problem 5: y = arcsin(x) + arccos(x)

Add the derivatives:

dy/dx = 1/√(1-x²) - 1/√(1-x²)

dy/dx = 0

This makes sense geometrically. arcsin(x) + arccos(x) = π/2 for valid x values. The derivative of a constant is zero.

Common Mistakes to Avoid

• Forgetting the chain rule. If the inside isn't just x, multiply by the derivative of the inside. Always.
• Dropping the negative sign on arccos, arccot, arccsc. These co-functions have negative derivatives. Don't forget that.
• Mixing up arcsin and arctan denominators. arcsin has √(1-x²). arctan has (1+x²). Different structures.
• Ignoring absolute values in arcsec and arccsc. The |x| in the denominator exists for a reason—it handles the symmetry of these functions.
• Forgetting domain restrictions. The derivative formula assumes the input is in the valid domain. Outside that, the function isn't defined.

Quick Reference Sheet

The "u" represents any differentiable function of x. Apply the chain rule by multiplying by u'.

When You'll Actually Use This

Inverse trig derivatives show up in integration. When you integrate 1/(1+x²), you get arctan(x). When you integrate 1/√(1-x²), you get arcsin(x). The derivatives are the reverse of the integrals.

They're also necessary when finding tangent lines to curves involving inverse trig functions, related rates problems, and optimization problems with trig inversions.

Know these formulas cold. The practice problems above should take you 30-45 minutes to work through completely. Do that, and you'll have them down.