From the free course AP Calculus AB on Lykke
3.4 Differentiating Inverse Trigonometric Functions
Key concepts: Differentiating inverse trigonometric functions · Unit 3: Differentiation—composite, implicit, and inverse functions · Checking the accuracy and appropriateness of solutions · Applying procedures with and without technology · Differentiating compositions correctly · Identifying omitted or incorrectly differentiated composition components · Verifying that factors in an answer come from the original problem
The derivatives of $\arcsin x$, $\arccos x$ and $\arctan x$, where they come from, how to combine them with the chain rule, and how to check an answer's sign, inner factor and domain.
3.4 Differentiating Inverse Trigonometric Functions
Topic 3.3 showed that the slope of an inverse function is the reciprocal of the original function's slope at the corresponding point. Applying that idea to sine, cosine and tangent gives derivative formulas for $\arcsin x$, $\arccos x$ and $\arctan x$ that, surprisingly, contain no trigonometric functions at all.
Differentiating inverse trigonometric functions
The inverse trigonometric functions are restricted inverses of trigonometric functions. Restriction is essential because $\sin x$, $\cos x$, and $\tan x$ repeat values and therefore are not one-to-one on all real numbers.
On their standard domains and ranges, the derivative rules are:
| Function | Derivative |
|---|---|
| $y=\arcsin x$ | $\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$ |
| $y=\arccos x$ | $\displaystyle \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}$ |
| $y=\arctan x$ | $\displaystyle \frac{dy}{dx}=\frac{1}{1+x^2}$ |
For a composition, the derivative includes the derivative of the inside expression. For example, with $y=\arctan(3x^2)$,
$$\frac{dy}{dx} =\frac{1}{1+(3x^2)^2}\cdot 6x =\frac{6x}{1+9x^4}.$$
The factor $6x$ is not optional: it records how quickly the input $3x^2$ changes.
A geometric check for $\arcsin x$
Let $y=\arcsin x$. Then $\sin y=x$, with $-\frac{\pi}{2}\le y\le\frac{\pi}{2}$. Differentiating gives
$$\cos y\frac{dy}{dx}=1,$$
so
$$\frac{dy}{dx}=\frac{1}{\cos y}.$$
Because $\cos y\ge0$ on this restricted range and $\sin y=x$,
$$\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2},$$
therefore
$$\frac{d}{dx}(\arcsin x)=\frac{1}{\sqrt{1-x^2}}.$$
The square root is required. Replacing it with $\cos^{-1}x$ would confuse an inverse function with a reciprocal and would produce the wrong expression.
Accuracy and appropriateness checks
Topic 3.3 and Topic 3.4 explicitly assess 3.G Confirm that solutions are accurate and appropriate. A reliable check asks three questions:
- Does the answer contain only factors from the original function and its differentiation rules?
- Was every composition differentiated, including the inside expression?
- Does the result fit the domain and sign of the function?
For instance, a proposed derivative of $\arccos(2x)$ equal to $\frac{1}{\sqrt{1-4x^2}}$ is incorrect because the inner derivative $2$ was skipped and the sign is wrong. The correct result is
$$-\frac{2}{\sqrt{1-4x^2}}.$$
A proposed answer with an unexplained factor such as $7$ should also be rejected: that factor did not arise from the original expression.
Misconception check: $\arctan x$ is not $\frac{1}{\tan x}$, and $\arcsin x$ is not $\frac{1}{\sin x}$. The prefix “arc” identifies an inverse function.
The procedure represented by 1.E Apply appropriate mathematical rules or procedures, with and without technology requires both symbolic work and technology-supported verification. A graphing calculator can compare a numerical derivative or graph slope, but it does not replace identifying the correct inverse, domain, sign, or missing factor.
Retrieval check
Find $\dfrac{d}{dx}\left[\arcsin(5x)\right]$ and explain one check that supports your answer.
The result is
$$\frac{5}{\sqrt{1-25x^2}}.$$
The factor $5$ comes from differentiating the inside expression $5x$, and the denominator requires $|5x|\le1$, confirming that the formula is appropriate on the real domain.