Finding the Derivative of sin−1x

Finding the derivative of a function is a fundamental concept in calculus. We are asked to find the derivative of the inverse sine function, $\sin^{-1}x$, with respect to $x$. This function is also sometimes written as $\arcsin(x)$. Understanding how to find the derivative of sin⁻¹x is important for solving many problems in calculus.

Understanding Inverse Trigonometric Functions

The function $y = \sin^{-1}x$ is the inverse of the sine function. It means that $y$ is the angle whose sine is $x$. For this function to be well-defined, we restrict the domain of the sine function to $[-\frac{\pi}{2}, \frac{\pi}{2}]$, and its range is $[-1, 1]$. Consequently, the domain of $\sin^{-1}x$ is $[-1, 1]$, and its range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Finding the derivative of sin⁻¹x falls under the topic of differentiation of inverse trigonometric functions.

Step-by-Step Differentiation of sin−1x

Let the given function be $y = \sin^{-1}x$. To find its derivative, $\frac{dy}{dx}$, we can use the technique of implicit differentiation.

1. Start with the function: $y = \sin^{-1}x$
2. Rewrite this equation by taking the sine of both sides: $\sin y = x$
3. Now, differentiate both sides of the equation $\sin y = x$ with respect to $x$. Remember that $y$ is a function of $x$, so we use the chain rule when differentiating $\sin y$.
4. Differentiating $\sin y$ with respect to $x$: $\frac{d}{dx}(\sin y) = \cos y \cdot \frac{dy}{dx}$
5. Differentiating $x$ with respect to $x$: $\frac{d}{dx}(x) = 1$
6. Equating the derivatives: $\cos y \cdot \frac{dy}{dx} = 1$
7. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{\cos y}$

Now we need to express $\cos y$ in terms of $x$. We know from the basic trigonometric identity that $\sin^2 y + \cos^2 y = 1$.

Since $\sin y = x$, we can substitute $x$ into the identity:

$x^2 + \cos^2 y = 1$

Solving for $\cos^2 y$:

$\cos^2 y = 1 - x^2$

Taking the square root of both sides:

$\cos y = \pm \sqrt{1 - x^2}$

However, we need to consider the range of $y$ for $y = \sin^{-1}x$. The range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. In this interval, the cosine function $\cos y$ is non-negative ($\cos y \ge 0$). Therefore, we must choose the positive square root:

$\cos y = \sqrt{1 - x^2}$

Substitute this expression for $\cos y$ back into the equation for $\frac{dy}{dx}$:

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

This gives us the derivative of sin⁻¹x with respect to $x$. This is a standard result in calculus for the derivative of inverse trigonometric functions.

Verifying the Options

Let's compare our result with the given options:

1. $\rm\frac{1}{\sqrt{1−x^2}}$
2. $\rm\frac{1}{\sqrt{x^2−1}}$
3. $\rm\frac{1}{\sqrt{1+x^2}}$
4. none of these

Our calculated derivative of sin⁻¹x is $\frac{1}{\sqrt{1 - x^2}}$, which matches the first option.

Summary of the Derivative of sin−1x

To summarize the differentiation process for finding the derivative of sin⁻¹x:

• Start with $y = \sin^{-1}x$.
• Rewrite as $x = \sin y$.
• Differentiate implicitly with respect to $x$: $1 = \cos y \frac{dy}{dx}$.
• Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{\cos y}$.
• Use the identity $\cos y = \sqrt{1 - \sin^2 y}$ and substitute $\sin y = x$.
• This leads to $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$.

This fundamental result is frequently used in higher mathematics and various applications requiring calculus.