3.4 - Differentiating Inverse Trigonometric Functions

Notation

Inverse trig functions can be written two equivalent ways:

\sin^{-1}(x) = \arcsin(x)

\cos^{-1}(x) = \arccos(x)

\tan^{-1}(x) = \arctan(x)

Again,

\sin^{-1}(x)

does not mean

\frac{1}{\sin x}

. It means the inverse function of sine.

The Three Primary Formulas

These are organized by pattern, which makes memorizing them easier.

Sine inverse:

\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1 - x^2}}

Secant inverse:

\frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{|x|\sqrt{x^2 - 1}}

Tangent inverse:

\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1 + x^2}

Memorization Strategy

Rule 1: “S” means Subtraction and Square root. Functions starting with S (sine, secant) have subtraction in the expression and a square root.

Rule 2: the “i” in sine. The “i” in sine looks like a “1,” so the 1 comes first:

1 - x^2

. Secant has no “i,” so the 1 comes last:

x^2 - 1

Rule 3: secant’s “C” stands for “Oh Crap!” The “c” in secant reminds you: “oh crap, there’s an absolute value!” That gives you

|x|

Rule 4: tangent has no “S.” No subtraction, no square root. Just addition:

1 + x^2

The Three Complementary Formulas

Functions starting with “co” are just the negatives of their counterparts.

Cosine inverse:

\frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1 - x^2}}

Cosecant inverse:

\frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{|x|\sqrt{x^2 - 1}}

Cotangent inverse:

\frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{1 + x^2}

Pattern: cosine inverse is the negative of sine inverse, and so on down the line.

Examples with the Chain Rule

Example 1: sine inverse

Find

\frac{d}{dx}[\sin^{-1}(3x)]

Starts with S, so square root and subtraction, with the 1 first:

\frac{d}{dx}[\sin^{-1}(3x)] = \frac{1}{\sqrt{1 - (3x)^2}} \cdot 3 = \frac{3}{\sqrt{1 - 9x^2}}

Common error: do NOT simplify

\sqrt{1 - 9x^2}

1 - 3x

. The square root applies to the entire expression

(1 - 9x^2)

, not to each term separately.

Example 2: tangent inverse

Find

\frac{d}{dx}[\tan^{-1}(2x^2)]

No “S,” so no square root, just addition:

\frac{d}{dx}[\tan^{-1}(2x^2)] = \frac{1}{1 + (2x^2)^2} \cdot 4x = \frac{4x}{1 + 4x^4}

Common error: don’t try to “cancel” the 4 and the

x

\text{WRONG: } \frac{4x}{1 + 4x^4} \neq \frac{1}{1 + x^3}

You can’t split a denominator like that. The whole expression

1 + 4x^4

is grouped together.

Example 3: secant inverse

Find

\frac{d}{dx}[\sec^{-1}(5x)]

Starts with S, so square root and subtraction (1 comes last). Has a C, so absolute value around

x

\frac{d}{dx}[\sec^{-1}(5x)] = \frac{1}{|5x|\sqrt{(5x)^2 - 1}} \cdot 5 = \frac{5}{|5x|\sqrt{25x^2 - 1}}

Cancel the 5’s:

= \frac{1}{|x|\sqrt{25x^2 - 1}}

Simplifying Secant Inverse Derivatives

These need careful handling of absolute values.

Strategy: check the sign before simplifying

Test with a negative number (like $x = -1$ ) to see if the result should be positive or negative

Simplify algebraically

Decide whether the final answer needs an absolute value

Example

Simplify

\dfrac{9x^2}{3|x|\sqrt{9x^6 - 1}}

Test with

x = -1

\frac{9(-1)^2}{3|-1|\sqrt{9(-1)^6 - 1}} = \frac{9}{3(1)\sqrt{8}} = \frac{3}{\sqrt{8}} > 0

The result is positive even when we plug in a negative number.

Now simplify:

\frac{9x^2}{3|x|\sqrt{9x^6 - 1}} = \frac{3x^2}{|x|\sqrt{9x^6 - 1}}

The useful fact here is that

\frac{x^2}{|x|} = |x|

, since

x^2 = |x|^2

. So:

= \frac{3|x|}{\sqrt{9x^6 - 1}}

Since we determined the answer has to be positive, and

|x|

guarantees that, we keep the absolute value.

Another example

Simplify

\dfrac{4x^2}{2|x|\sqrt{4x^2 - 1}}

Test with

x = -1

\frac{4(-1)^2}{2|-1|\sqrt{4(-1)^2 - 1}} = \frac{4}{2\sqrt{3}} > 0

Positive, good. Now simplify:

\frac{4x^2}{2|x|\sqrt{4x^2 - 1}} = \frac{2x^2}{|x|\sqrt{4x^2 - 1}}

Check the

\frac{x^2}{|x|}

piece both ways: - If

x < 0

\frac{x^2}{|x|} = \frac{x^2}{-x} = -x

, which is positive - If

x > 0

\frac{x^2}{|x|} = \frac{x^2}{x} = x

, which is positive

Either way

\frac{x^2}{|x|} = |x|

, so:

= \frac{2|x|}{\sqrt{4x^2 - 1}}

The key is to test whether the original expression comes out positive or negative, then make sure your simplified form behaves the same way.

Domain and Range of Inverse Trig Functions

These matter a lot when you’re evaluating inverse trig expressions.

Arcsine: $y = \sin^{-1}(x)$

Domain:

-1 \leq x \leq 1

Range:

-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

3.4 - Differentiating Inverse Trigonometric Functions

Arccosine: $y = \cos^{-1}(x)$

Domain:

-1 \leq x \leq 1

Range:

0 \leq y \leq \pi

Arctangent: $y = \tan^{-1}(x)$

Domain:

-\infty < x < \infty

(all real numbers)

Range:

-\frac{\pi}{2} < y < \frac{\pi}{2}

(open interval)

Why an open interval? Because

\tan \theta = \frac{\sin \theta}{\cos \theta}

, and

\cos(\pm\frac{\pi}{2}) = 0

, which means dividing by zero.

Evaluating Inverse Trig Functions

Example 1: evaluate

\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)

Translation: what angle has a sine value of

\frac{\sqrt{3}}{2}

From the unit circle,

\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}

, and

\frac{\pi}{3}

is in the range

[-\frac{\pi}{2}, \frac{\pi}{2}]

\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}

Example 2: evaluate

\cos^{-1}\left(-\frac{1}{2}\right)

Translation: what angle has a cosine value of

-\frac{1}{2}

From the unit circle,

\cos(\frac{2\pi}{3}) = -\frac{1}{2}

, and

\frac{2\pi}{3}

is in the range

[0, \pi]

\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}

Example 3: evaluate

\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)

Translation: what angle has a tangent value of

\frac{1}{\sqrt{3}}

Recall

\tan \theta = \frac{\sin \theta}{\cos \theta}

. At

\theta = \frac{\pi}{6}

, we have

\sin(\frac{\pi}{6}) = \frac{1}{2}

and

\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}

, so:

\tan\left(\frac{\pi}{6}\right) = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}

Since

\frac{\pi}{6}

is in the range

(-\frac{\pi}{2}, \frac{\pi}{2})

\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}

Where the Formulas Come From

You don’t have to take these formulas on faith. Each one comes from the chain rule plus the definition of an inverse function, the same way as the formula in 3.3.

Arcsine. Let

y = \sin^{-1}(x)

. By definition, that means

\sin(y) = x

Differentiate both sides with respect to

x

. The left side needs the chain rule, just like implicit differentiation:

\cos(y) \cdot \frac{dy}{dx} = 1 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{\cos(y)}

Now write

\cos(y)

in terms of

x

. From

\sin^2(y) + \cos^2(y) = 1

, we get

\cos(y) = \pm\sqrt{1 - \sin^2(y)} = \pm\sqrt{1 - x^2}

. The range of arcsine is

-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}

, and cosine is never negative there, so we keep the positive root:

\frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1 - x^2}}

This is one reason the range matters: it decides the sign. Also, at

x = \pm 1

we get

\cos(y) = 0

, so the derivative doesn’t exist at the ends of the domain.

Arccosine works the same way. Starting from

\cos(y) = x

gives

-\sin(y) \cdot \frac{dy}{dx} = 1

, and that minus sign from the derivative of cosine is why the “co” formulas are negative.

Arctangent. Let

y = \tan^{-1}(x)

, so

\tan(y) = x

. Differentiate both sides:

\sec^2(y) \cdot \frac{dy}{dx} = 1 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{1}{\sec^2(y)}

Use the identity

\sec^2(y) = 1 + \tan^2(y) = 1 + x^2

\frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1 + x^2}

Using the inverse function formula instead. The formula from 3.3 works at a single point too. Find the derivative of

\sin^{-1}(x)

x = \frac{1}{2}

. Here

f(x) = \sin x

(restricted to

-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}

), so

f'(x) = \cos x

, and

f^{-1}\left(\frac{1}{2}\right) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}

(f^{-1})'\left(\frac{1}{2}\right) = \frac{1}{f'\left(\frac{\pi}{6}\right)} = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}

Check with the arcsine formula:

\frac{1}{\sqrt{1 - \left(\frac{1}{2}\right)^2}} = \frac{1}{\sqrt{\frac{3}{4}}} = \frac{2}{\sqrt{3}}

✓

Key Takeaways

Notation: $\sin^{-1}(x) = \arcsin(x)$ , an inverse function, not a reciprocal

“S” rule: functions starting with S (sine, secant) have subtraction and a square root

“i” in sine: the 1 comes first: $\sqrt{1 - x^2}$

“C” in secant: “oh crap!” means add an absolute value: $\frac{1}{|x|\sqrt{x^2 - 1}}$

“Co” means negative: cosine, cosecant, and cotangent derivatives are the negatives of their counterparts

Tangent is simple: no S, so no subtraction and no square root: $\frac{1}{1 + x^2}$

Chain rule always applies: multiply by the derivative of the inside function

Where they come from: rewrite $y = \sin^{-1}(x)$ as $\sin(y) = x$ and differentiate implicitly, or use $\frac{1}{f'(f^{-1}(x))}$ from 3.3

Don’t split denominators: $\sqrt{1 - x^2}$ and $1 + x^2$ are single grouped expressions

Domain matters: arcsine and arccosine only accept inputs in $[-1, 1]$

Range matters: know which quadrants each inverse function outputs to

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