Question

What is the slope of the line tangent to the graph of \(y=\sin ^{-1} x\) at \(x=0 ?\)

Short answer

Answer: The slope of the tangent line at \(x=0\) is \(1\).

Step-by-step solution

Find the derivative of \(y=\sin^{-1} x\)

First, let's rewrite the equation as \(\sin y = x\). Now, we can differentiate both sides with respect to \(x\). On the left side, use the chain rule where we have the derivative of \(\sin y\) times \(dy/dx\). $$\frac{d}{dx}\sin y=\frac{d}{dx}(x)$$ $$\cos y \frac{dy}{dx} = 1$$

Solve for \(\frac{dy}{dx}\)

Isolate \(\frac{dy}{dx}\) by dividing both sides by \(\cos y\): $$\frac{dy}{dx} = \frac{1}{\cos y}$$

Substitute the original function in place of y to get the derivative of the function \(y=\sin^{-1} x\) in terms of x.

We have \(y=\sin^{-1} x\). Therefore, \(y=\sin^{-1}(x)\). Substitute this back into the derivative: $$\frac{dy}{dx} = \frac{1}{\cos (\sin^{-1} x)}$$

Simplify the expression

Use the trigonometric identity \(\cos^2 x + \sin^2 x = 1\) to get $$\cos(\sin^{-1} x) = \sqrt{1 - \sin^2(\sin^{-1} x)}$$ On the right side, we have \(\sin(\sin^{-1} x) = x\). Therefore, $$\cos(\sin^{-1} x) = \sqrt{1 - x^2}$$ Now, substitute this back into the derivative we found in Step 3: $$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$$

Find the slope of the tangent line at \(x=0\)

Plug in \(x=0\) to find the slope of the tangent line at that point: $$\frac{dy}{dx}(0) = \frac{1}{\sqrt{1 - 0^2}} = \frac{1}{\sqrt{1}} = 1$$ The slope of the line tangent to the graph of \(y=\sin^{-1} x\) at \(x=0\) is \(1\).

Key concepts

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find an angle when we know the ratios of the sides of a right-angled triangle. For any given ratio, the corresponding inverse trigonometric function will give us an angle that produces that ratio when applied to a certain trigonometric function.

For example, the inverse sine function, written as \( \text{sin}^{-1} \) or \( \text{asin} \), returns an angle whose sine is the given number. Similarly, \( \text{cos}^{-1} \) and \( \text{tan}^{-1} \) give us angles whose cosine and tangent are the given numbers, respectively.

These functions are crucial in trigonometry, calculus, and many applied fields because they let us work backwards from known values to find unknown angles. Functions like \( \text{sin}^{-1} \) are defined only for inputs between -1 and 1, since sine values are restricted to this range.

Derivative of Inverse Sine

The derivative of the inverse sine function, \( \text{sin}^{-1} \), is particularly interesting. To understand this, we recall that the derivative of a function at a certain point gives us the slope of the tangent line to the graph of the function at that point.

To find the derivative of \( y = \text{sin}^{-1}(x) \), we use the implicit differentiation method, where we first write it as \( \text{sin}(y) = x \) and then differentiate both sides with respect to \( x \). As seen in the solution, the derivative of \( y = \text{sin}^{-1}(x) \) with respect to \( x \) is \( \frac{dy}{dx} = \frac{1}{\(\sqrt{1 - x^2}\)} \), provided that \( x \) is in the interval \( [-1,1] \).

At \( x = 0 \), the slope of the tangent line \( \frac{dy}{dx} \) becomes 1, meaning the tangent line is at a 45-degree angle to the horizontal, a moderately steep incline.

Chain Rule

In calculus, the chain rule is used to differentiate compositions of functions. It tells us how to find the derivative of a composite function by taking the derivative of the outer function and multiplying it by the derivative of the inner function.

The general formula for the chain rule is: \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x) \]

When applying the chain rule to inverse trigonometric functions where the composition is not explicit, we often have to use implicit differentiation, as we did in the solution for the slope of the tangent line to \( y = \text{sin}^{-1}(x) \). This is because the composition of functions is hidden in the inverse function notation. By using implicit differentiation combined with the chain rule, we arrive at the correct expression for the derivative.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value within their domains. One of the most fundamental identities is the Pythagorean identity: \( \text{cos}^2(x) + \text{sin}^2(x) = 1 \), which is derived from the Pythagorean theorem and defines a fundamental relationship between the sine and cosine of an angle.

In the problem's solution, the Pythagorean identity is used to simplify \( \text{cos}(\text{sin}^{-1}(x)) \) into \( \sqrt{1 - x^2} \). By understanding and applying these trigonometric identities, we can manipulate and simplify expressions involving trigonometric functions, often making the difference between a problem that is solvable and one that seems intractable.