Problem 55 - University Calculus: Early Transcendentals

Question

How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. Derivative of arcsine Assume that \(y=\sin ^{-1} x\) is a differentiable function of \(x\). By differentiating the equation \(x=\sin y\) implicitly, show that \(d y / d x=1 / \sqrt{1-x^{2}}\)

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\); \(\frac{dy}{dx}\) and \(\frac{dx}{dy}\) are inverses, illustrating the slope change.

Step 1: Understand the Given Equation

You're given the equation \(y = \sin^{-1}(x)\), which implies \(x = \sin(y)\). We need to differentiate this relationship to find \( \frac{dy}{dx} \).

Step 2: Apply Implicit Differentiation

Differentiate the equation \(x = \sin(y)\) with respect to \(x\). This involves differentiating both sides.The derivative of \(x\) with respect to \(x\) is 1, and the derivative of \(\sin(y)\) with respect to \(x\) is \(\cos(y) \frac{dy}{dx}\), applying the chain rule.

Step 3: Set Up the Differential Equation

From the derivatives, we have:\[1 = \cos(y) \frac{dy}{dx}\]This equation relates \( \frac{dy}{dx} \) to \( \cos(y) \).

Step 4: Solve for \(\frac{dy}{dx}\)

Rearrange the equation to solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{1}{\cos(y)}\]

Step 5: Relate \(\cos(y)\) to \(x\)

Recall that since \(x = \sin(y)\), by the identity \(\sin^2(y) + \cos^2(y) = 1\), we have:\[\cos^2(y) = 1 - \sin^2(y) = 1 - x^2\]Thus, \( \cos(y) = \sqrt{1 - x^2} \) if \(y\) is in the range of \(\sin^{-1}(x)\), where \(\cos(y) > 0\).

Step 6: Result for \(\frac{dy}{dx}\)

Substitute into the expression for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\]

Step 7: Explore the Relationship Between \(\frac{dy}{dx}\) and \(\frac{dx}{dy}\)

Geometrically, \(\frac{dy}{dx}\) is the slope of the tangent line at a point on the curve \(y = \sin^{-1}(x)\), while \(\frac{dx}{dy}\) would be \(\cos(y)\), the reciprocal of \(\frac{dy}{dx}\). They are multiplicative inverses: \(\frac{dy}{dx} \cdot \frac{dx}{dy} = 1\).

Step 8: Conclude the Geometric Relationship

The geometric interpretation is that the tangent to \(y = \sin^{-1}(x)\) at any point has a slope \(\frac{dy}{dx}\), and the arc has a supplementary angle described by \(\frac{dx}{dy}\) as the cotangent. This reflects how the arc length changes in relation to the vertical and horizontal changes.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used for finding derivatives of functions that are not easily expressed as a simple formula. Instead, these functions are often described by an equation involving multiple variables.

When you differentiate both sides of an equation with respect to one variable while considering another variable as a function of the first, that's implicit differentiation.

• Start with a relation like \(x = \sin(y)\).
• Differentiating both sides with respect to \(x\) gives you: \[ \frac{d}{dx}(x) = \frac{d}{dx}(\sin(y)) \] Simplifying, we have: \[ 1 = \cos(y) \frac{dy}{dx} \] Applying the chain rule is crucial here because \(y\) is implicitly defined.
• Notice that \(\frac{dy}{dx}\) is "buried" in the derivative of the right-hand side and needs to be isolated by rearranging the equation.

Geometric Interpretation of Derivatives

Derivatives represent the rate of change or the slope of a curve at a given point. On a graph, this is visually interpreted as the slope of the tangent line to the curve at that point.

In the case of inverse trigonometric functions, like \(y = \sin^{-1}(x)\), the geometric interpretation is fascinating:

• \(\frac{dy}{dx}\): This derivative tells us the slope of the tangent line at any given point on the graph of \(y = \sin^{-1}(x)\). The slope represents how steeply the curve climbs or falls as \(x\) changes.
• \(\frac{dx}{dy}\): This value represents the slope of the tangent when viewed inversely, offering a mirror image or reciprocal perspective of change.

These two derivatives are multiplicative inverses:

• The multiplicative inverse relationship: \(\frac{dy}{dx} \cdot \frac{dx}{dy} = 1\). This reflects that for every angle's change in the arc, there's a corresponding horizontal shift portrayed through these derivatives.
• The tangent line on the circle or arc described by the function has these directional changes matching these inverted slopes, showing how tightly interconnected change in "height" and "width" really are on a curve.

Inverse Trigonometric Functions

Inverse trigonometric functions are the "reverse" of the usual trigonometric functions like sine and cosine. They work to find angles when you know the trigonometric value.

For example, \(\sin^{-1}(x)\) (or arcsine) tells us what angle \(y\) would have a sine value of \(x\). It's defined only for \(-1 \leq x \leq 1\) and results in angles of \(-\pi/2 \leq y \leq \pi/2\).

• These functions are very useful in geometry and trigonometry, providing means to translate values back to angles.
• When differentiating \(y = \sin^{-1}(x)\), we use implicit differentiation because \(y\) is not immediately solved for, as it's nested in the expression \(x = \sin(y)\).
• The resulting derivative, \(\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\), shows how tiny changes in \(x\) affect changes in \(y\) through this function, though steep as \(x\) moves towards \(-1\) or \(1\).

This process reveals how inverse trigonometric functions are differentiable and express change more dynamically than initially meets the eye. The range and values significantly impact derivative characteristics.