Derivatives of Inverse Trigonometric Functions: Videos & Practice Problems

Understanding the derivatives of inverse trigonometric functions is essential for calculus, particularly when working with functions involving inverse sine and inverse cosine. To derive the formula for the derivative of the inverse sine function, we start with the relationship defined by the function itself. If \( y = \sin^{-1}(x) \), then it follows that \( x = \sin(y) \). By applying implicit differentiation, we differentiate both sides with respect to \( x \).

The derivative of \( x \) with respect to \( x \) is simply 1. For the right side, using the chain rule, the derivative of \( \sin(y) \) is \( \cos(y) \cdot \frac{dy}{dx} \). This leads us to the equation:

\[1 = \cos(y) \cdot \frac{dy}{dx}\]

Rearranging gives us:

\[\frac{dy}{dx} = \frac{1}{\cos(y)}\]

To express this in terms of \( x \), we utilize the Pythagorean identity \( \sin^2(y) + \cos^2(y) = 1 \). Solving for \( \cos(y) \) yields:

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

Substituting this back into our derivative expression results in:

\[\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}\]

This formula is valid under the condition that \( |x| < 1 \), reflecting the range of the inverse sine function. Consequently, the derivative of the inverse sine function is:

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

Next, we can apply this knowledge to find the derivative of a composite function, such as \( f(x) = \sin^{-1}(3x + 2) \). Using the chain rule, we differentiate as follows:

\[f'(x) = \frac{1}{\sqrt{1 - (3x + 2)^2}} \cdot \frac{d}{dx}(3x + 2) = \frac{3}{\sqrt{1 - (3x + 2)^2}}\]

Similarly, the derivative of the inverse cosine function can be derived using a parallel process. If \( y = \cos^{-1}(x) \), then \( x = \cos(y) \). Following the same implicit differentiation steps, we find:

\[\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}\]

This derivative also holds under the condition \( |x| < 1 \). Thus, the derivative of the inverse cosine function is:

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

For example, to find the derivative of \( g(x) = 4 \cos^{-1}(6x) \), we apply the chain rule:

\[g'(x) = 4 \left(-\frac{1}{\sqrt{1 - (6x)^2}} \cdot 6\right) = -\frac{24}{\sqrt{1 - 36x^2}}\]

In summary, mastering the derivatives of inverse trigonometric functions is crucial for solving complex calculus problems. The derivatives are:

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

These derivatives will frequently appear in various calculus applications, making it beneficial to memorize them while also understanding the derivation process through implicit differentiation.

To find the derivative \(\frac{dy}{dx}\) at the ordered pair \((0, 1/2)\) for the equation \(4 \sin^{-1}(x) + \cos^{-1}(y) = \frac{\pi}{3}\), we will use implicit differentiation. We start by differentiating both sides of the equation with respect to \(x\).

The left side becomes:

\(\frac{d}{dx}[4 \sin^{-1}(x)] + \frac{d}{dx}[\cos^{-1}(y)]\)

Using the derivative formulas, we have:

\(\frac{d}{dx}[4 \sin^{-1}(x)] = \frac{4}{\sqrt{1 - x^2}}\)

For \(\cos^{-1}(y)\), we apply the chain rule:

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

Setting the derivative of the right side, which is a constant \(\frac{\pi}{3}\), to zero, we have:

\(\frac{4}{\sqrt{1 - x^2}} - \frac{1}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = 0\)

Rearranging gives:

\(-\frac{1}{\sqrt{1 - y^2}} \cdot \frac{dy}{dx} = -\frac{4}{\sqrt{1 - x^2}}\

Multiplying both sides by \(-\sqrt{1 - y^2}\) leads to:

\(\frac{dy}{dx} = \frac{4 \sqrt{1 - y^2}}{\sqrt{1 - x^2}}\)

Now, substituting the values \(x = \frac{1}{2}\) and \(y = 0\) into the equation:

\(\frac{dy}{dx} = \frac{4 \sqrt{1 - 0^2}}{\sqrt{1 - \left(\frac{1}{2}\right)^2}}\)

This simplifies to:

\(\frac{dy}{dx} = \frac{4 \cdot 1}{\sqrt{1 - \frac{1}{4}}} = \frac{4}{\sqrt{\frac{3}{4}}} = \frac{4}{\frac{\sqrt{3}}{2}} = \frac{4 \cdot 2}{\sqrt{3}} = \frac{8}{\sqrt{3}}\)

To express this in a more standard form, we can rationalize the denominator:

\(\frac{8}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{8\sqrt{3}}{3}\)

Thus, the final result for \(\frac{dy}{dx}\) at the ordered pair \((0, 1/2)\) is:

\(\frac{8\sqrt{3}}{3}\)

Understanding the derivatives of inverse trigonometric functions is essential for calculus, particularly when applying rules like implicit differentiation. The primary inverse trigonometric functions include inverse sine, inverse cosine, inverse tangent, inverse cotangent, inverse secant, and inverse cosecant. Each of these functions has specific derivative rules that can be derived through implicit differentiation.

For example, when finding the derivative of a function such as f(x) = 4x² * arctan(x), the product rule is applied. The product rule states that the derivative of a product of two functions is given by f'(x) = u'v + uv', where u and v are the two functions. In this case, the derivative of the inverse tangent function, arctan(x), is 1 / (1 + x²). Thus, the derivative can be expressed as:

f'(x) = 4x² * (1 / (1 + x²)) + arctan(x) * 8x

This simplifies to:

f'(x) = (4x² / (1 + x²)) + 8x * arctan(x)

In another example, consider the function g(x) = 6 * arccot(x) + 4 * arcsec(3x + 1). The derivative involves applying the derivative rules for both inverse cotangent and inverse secant. The derivative of arccot(x) is -1 / (1 + x²), while the derivative of arcsec(x) is 1 / (|x| * √(x² - 1)). When applying the chain rule for the term arcsec(3x + 1), the derivative becomes:

g'(x) = 6 * (-1 / (1 + x²)) + 4 * (1 / (|3x + 1| * √((3x + 1)² - 1))) * 3

Finally, for a function like h(x) = (arccsc(x))³, the power rule is utilized alongside the derivative of the inverse cosecant function, which is -1 / (|x| * √(x² - 1)). The derivative is calculated as:

h'(x) = 3 * (arccsc(x))² * (-1 / (|x| * √(x² - 1)))

It is important to note that for the inverse secant and inverse cosecant functions, the condition |x| > 1 must be satisfied, reflecting the domains of these functions. Mastering these derivatives through practice will enhance your ability to tackle more complex calculus problems efficiently.