Derivatives Of Inverse Trigonometric Functions

What Are Inverse Trigonometric Functions?

Inverse trigonometric functions are functions that reverse, or “undo,” the standard trigonometric functions. For a given value of a trigonometric function, the inverse function tells you the angle (in radians) that produces that value. They are often denoted with “arc” (like arcsin) or with a superscript -1 (like sin⁻¹).

For example, we know that $\sin (\frac{\pi }{6})=\frac{1}{2}$. The inverse sine function, or arcsin, reverses this relationship: $\arcsin (\frac{1}{2})=\frac{\pi }{6}$. It answers the question, “What angle has a sine of $\frac{1}{2}$?”

However, there's a small complication. Since trigonometric functions are periodic (they repeat their values), there are infinitely many angles that have a sine of $\frac{1}{2}$ (e.g., $\frac{5\pi }{6},\frac{13\pi }{6}$, etc.). To make the inverse a true function (meaning it gives only one output for each input), we must restrict its range of possible output angles. Each inverse trigonometric function has a specific, standard range.

Understanding these ranges is crucial, as they ensure our derivative formulas are well-defined.

How Do We Find the Derivative of an Inverse Function?

Before we tackle the inverse trig functions specifically, let's establish a general method for finding the derivative of any inverse function. This method relies on a clever technique called implicit differentiation.

Let's say we have a function $y=f^{-1}(x)$. By the definition of an inverse function, this is equivalent to saying $x=f(y)$.

Now, let's differentiate both sides of the equation $x=f(y)$ with respect to $x$:

The left side is simple: $\frac{d}{dx}(x)=1$.

For the right side, we must use the chain rule because $y$ is a function of $x$. The derivative of $f(y)$ with respect to $y$ is $f^{\prime }(y)$, and the derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$. So, the chain rule gives us:

Putting it all together, we have:

Our goal is to find $\frac{dy}{dx}$, so we just solve for it algebraically:

Since we started with $y=f^{-1}(x)$, we can substitute this back in to get our final, general formula:

This powerful formula tells us that the derivative of an inverse function at a point $x$ is the reciprocal of the derivative of the original function evaluated at $f^{-1}(x)$. We will use this exact strategy to derive the formulas for our inverse trigonometric functions.

How Do You Find the Derivative of Arcsin(x)?

Let's use the method from the previous section to find the derivative of $y=\arcsin (x)$. This is arguably the most common inverse trigonometric derivative you'll encounter.

1. Rewrite the function: If $y=\arcsin (x)$, then the equivalent statement is $x=\sin (y)$. This is valid for $x$ in $[-1,1]$ and $y$ in $[-\frac{\pi }{2},\frac{\pi }{2}]$.
2. Differentiate implicitly: We differentiate both sides of $x=\sin (y)$ with respect to $x$. $$\frac{d}{dx}(x)=\frac{d}{dx}(\sin (y))$$ $$1=\cos (y)\cdot \frac{dy}{dx}$$
3. Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{1}{\cos (y)}$$
4. Express in terms of x: Our derivative is in terms of $y$, but we want it in terms of $x$. We need to find a way to write $\cos (y)$ using $x$. We can use the Pythagorean identity: ${\sin }^2(y)+{\cos }^2(y)=1$.

   Solving for $\cos (y)$, we get $\cos (y)=\pm \sqrt{1-{\sin }^2(y)}$. Since we know $x=\sin (y)$, we can substitute that in: $\cos (y)=\pm \sqrt{1-x^2}$.

   Which sign do we choose? Remember the range restriction for $y=\arcsin (x)$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$. In this interval (Quadrants I and IV), the cosine function is always non-negative. Therefore, we take the positive root: $\cos (y)=\sqrt{1-x^2}$.

5. Final substitution: Now substitute this back into our expression for $\frac{dy}{dx}$.

When combined with the chain rule, if $u$ is a function of $x$, the formula becomes:

What Is the Derivative of Arccos(x)?

The derivation for $y=\arccos (x)$ is very similar to that of arcsin, with one key difference in sign.

1. Rewrite the function: $y=\arccos (x)$ is equivalent to $x=\cos (y)$, for $y$ in $[0,\pi ]$.
2. Differentiate implicitly: Differentiate $x=\cos (y)$ with respect to $x$. Remember that the derivative of cosine is negative sine. $$\frac{d}{dx}(x)=\frac{d}{dx}(\cos (y))$$ $$1=-\sin (y)\cdot \frac{dy}{dx}$$
3. Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=-\frac{1}{\sin (y)}$$
4. Express in terms of x: We again use ${\sin }^2(y)+{\cos }^2(y)=1$. This gives $\sin (y)=\pm \sqrt{1-{\cos }^2(y)}$. Since $x=\cos (y)$, we have $\sin (y)=\pm \sqrt{1-x^2}$.

   For $y=\arccos (x)$, the range is $[0,\pi ]$. In this interval (Quadrants I and II), the sine function is always non-negative. So, we choose the positive root: $\sin (y)=\sqrt{1-x^2}$.

5. Final substitution: Substitute this into our expression for $\frac{dy}{dx}$.

Notice this is simply the negative of the derivative of $\arcsin (x)$! This relationship stems from the identity $\arcsin (x)+\arccos (x)=\frac{\pi }{2}$.

How Is the Derivative of Arctan(x) Found?

The derivative of $y=\arctan (x)$ is another essential formula to know. Its derivation uses a different Pythagorean identity.

1. Rewrite the function: If $y=\arctan (x)$, then $x=\tan (y)$. This is valid for all real numbers $x$ and for $y$ in $(-\frac{\pi }{2},\frac{\pi }{2})$.
2. Differentiate implicitly: Differentiate $x=\tan (y)$ with respect to $x$. The derivative of $\tan (y)$ is ${\sec }^2(y)$. $$\frac{d}{dx}(x)=\frac{d}{dx}(\tan (y))$$ $$1={\sec }^2(y)\cdot \frac{dy}{dx}$$
3. Solve for $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{1}{{\sec }^2(y)}$$
4. Express in terms of x: We need to relate ${\sec }^2(y)$ back to $x=\tan (y)$. We use the Pythagorean identity $1+{\tan }^2(y)={\sec }^2(y)$.

   Since $x=\tan (y)$, we can substitute directly: ${\sec }^2(y)=1+x^2$.

5. Final substitution: Put this result back into our derivative expression.

This result is particularly nice because it doesn't involve a square root. The chain rule version is:

What About the Other Three Inverse Trig Functions?

The remaining three inverse trigonometric functions—arccotangent, arcsecant, and arccosecant—can be derived using the same implicit differentiation method. We will present their formulas here. Notice the pattern: the derivative of each “co-” function is the negative of its corresponding function's derivative.

• Derivative of Arccotangent (arccot): The derivative of $\text{arccot}(x)$ is the negative of the $\arctan (x)$ derivative. $$\frac{d}{dx}(\text{arccot}(x))=-\frac{1}{1+x^2}$$
• Derivative of Arcsecant (arcsec): The derivation for arcsecant is a bit trickier due to an absolute value, which is needed to ensure the derivative's sign is correct for both positive and negative $x$. $$\frac{d}{dx}(\text{arcsec}(x))=\frac{1}{|x|\sqrt{x^2-1}}$$
• Derivative of Arccosecant (arccsc): Similarly, the derivative of $\text{arccsc}(x)$ is the negative of the $\text{arcsec}(x)$ derivative. $$\frac{d}{dx}(\text{arccsc}(x))=-\frac{1}{|x|\sqrt{x^2-1}}$$

Quick Summary and Reference Table

Memorizing these six derivatives is essential for success in calculus. The patterns can help you remember them. The three “co-” functions (arccos, arccot, arccsc) have derivatives that are the negatives of their counterparts (arcsin, arctan, arcsec). Here is a complete reference table.

Remember to always apply the Chain Rule when the argument of the function is more complex than just $x$. If you have $f(u(x))$, the derivative is $f^{\prime }(u(x))\cdot u^{\prime }(x)$.

What Are Some Common Mistakes to Avoid?

When working with these derivatives, students often make a few predictable errors. Being aware of them is the first step to avoiding them!

• Forgetting the Chain Rule: This is the most common mistake. If you are differentiating $\arctan (3x)$, the answer is not just $\frac{1}{1+(3x{)}^2}$. You must multiply by the derivative of the inside function, $3x$, which is $3$. The correct answer is $\frac{3}{1+9x^2}$.
• Sign Errors: It's easy to mix up the signs. Remember the pattern: the derivatives of the “co-” functions (cosine, cotangent, cosecant) are all negative.
• Algebraic Simplification Errors: Be careful when substituting into the formulas. For example, in $\arcsin (x^3)$, the term $u^2$ becomes $(x^3{)}^2=x^6$, not $x^9$ or $x^5$. Double-check your exponent rules and distribution.
• Confusing Notation: The notation ${\sin }^{-1}(x)$ means $\arcsin (x)$, not $\frac{1}{\sin (x)}$, which is $\csc (x)$. This is a very important distinction. The superscript -1 indicates an inverse function, not a reciprocal exponent, in this context.
• Ignoring the Absolute Value: Forgetting the absolute value in the derivatives of $\text{arcsec}(x)$ and $\text{arccsc}(x)$ is a frequent error. It's necessary to make the formula work for negative values of $x$.