Problem 66 - Calculus - AP Edition

Question

Find \(d y / d x\) by implicit differentiation. $$\sin ^{-1}(x y)=\cos ^{-1}(x-y)$$

Step-by-Step Solution

Answer

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

Step 1: Differentiate both sides

Differentiate both sides of the equation with respect to \(x\). The given equation is \(\sin^{-1}(xy) = \cos^{-1}(x-y)\). Remember to use the chain rule where necessary.

Step 2: Differentiate the left-hand side

For the left-hand side, differentiate \(\sin^{-1}(xy)\). The derivative of \(\sin^{-1}(u)\) with respect to \(x\) is \(\frac{1}{\sqrt{1-u^2}}\cdot\frac{du}{dx}\). Here, \(u = xy\), so \(\frac{du}{dx} = y + x\frac{dy}{dx}\). Therefore, the derivative of the left side is \(\frac{1}{\sqrt{1-(xy)^2}}(y + x\frac{dy}{dx})\).

Step 3: Differentiate the right-hand side

For the right-hand side, differentiate \(\cos^{-1}(x-y)\). The derivative of \(\cos^{-1}(v)\) with respect to \(x\) is \(-\frac{1}{\sqrt{1-v^2}}\cdot\frac{dv}{dx}\). Here, \(v = x-y\), so \(\frac{dv}{dx} = 1 - \frac{dy}{dx}\). Therefore, the derivative of the right side is \(-\frac{1}{\sqrt{1-(x-y)^2}}(1 - \frac{dy}{dx})\).

Step 4: Set derivatives equal and solve for \(\frac{dy}{dx}\)

Set the derivatives from Step 2 and Step 3 equal to one another: \[\frac{1}{\sqrt{1-(xy)^2}}(y + x\frac{dy}{dx}) = -\frac{1}{\sqrt{1-(x-y)^2}}(1 - \frac{dy}{dx})\]. Now, solve this equation for \(\frac{dy}{dx}\). Distribute and combine like terms to isolate \(\frac{dy}{dx}\).

Step 5: Rearrange and solve for \(\frac{dy}{dx}\)

After rearranging terms and simplifying, you obtain: \[y\sqrt{1-(x-y)^2} + \sqrt{1-(xy)^2} = \frac{dy}{dx} \left( x\sqrt{1-(x-y)^2} + \sqrt{1-(xy)^2} \right)\]. Solve for \(\frac{dy}{dx}\) by dividing both sides by \(x\sqrt{1-(x-y)^2} + \sqrt{1-(xy)^2}\).

Key Concepts

Chain Rule

The chain rule is a fundamental principle in calculus used for differentiating composite functions. In this exercise, both sides of the equation involve functions within functions, also known as nested functions, which need the chain rule for differentiation. The chain rule states: if a function is composed of two functions, say, \(f(g(x))\), the derivative is the product of the derivative of the outer function evaluated at the inner function and the derivative of the inner function. Formally, this is expressed as: \(\frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx}\).

When applying the chain rule to the left-hand side of our equation, \(\sin^{-1}(xy)\), we treat \(xy\) as an inner function. We first differentiate \(\sin^{-1}(u)\) with respect to \(u\), where \(u = xy\), giving us \(\frac{1}{\sqrt{1-u^2}}\). Then, multiply by the derivative of \(u\) with respect to \(x\), which results from applying the product rule to \(xy\): \(y + x\frac{dy}{dx}\).

Similarly, on the right-hand side, for \(\cos^{-1}(x-y)\), we use the chain rule. With the inner function \(v = x-y\), the derivative \(-\frac{1}{\sqrt{1-v^2}}\) is taken and multiplied by \(\frac{dv}{dx}\), resulting in \(1 - \frac{dy}{dx}\). This systematic approach through the chain rule allows the decomposition and differentiation of complex nested functions easily.

Trigonometric Functions

Trigonometric functions are pervasive in calculus due to their cyclic and oscillating nature. They describe angles and relate to the unit circle involving sine, cosine, and other related functions. This exercise uses both the inverse sine and cosine functions, which are represented as \(\sin^{-1}(x)\) and \(\cos^{-1}(x)\), respectively.

Each inverse trigonometric function has a standard derivative formula. For \(y = \sin^{-1}(x)\), the derivative is \(\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}\). This helps us understand the rate of change of the angle concerning \(x\). Similarly, \(y = \cos^{-1}(x)\) has a derivative of \(\frac{d}{dx}\cos^{-1}(x) = -\frac{1}{\sqrt{1-x^2}}\). The negative sign indicates that the inverse cosine decreases as \(x\) increases, contrasting with the behavior of the inverse sine.

This exercise highlights how the combination of these functions with other variables requires careful differentiation, leveraging their derivatives alongside the chain rule. Keeping these derivatives in mind is crucial when dealing with trigonometric functions within calculus problems.

Inverse Trigonometric Functions

Inverse trigonometric functions are used when given a ratio or coordinate, to find the angle that produces it. These functions are the inverses of conventional trigonometric functions, allowing for the conversion from decimal answers back to angles.

The problem involves both \(\sin^{-1}\) and \(\cos^{-1}\). It’s essential to recall that their range is restricted: \(\sin^{-1}(x)\) values range between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), whereas \(\cos^{-1}(x)\) covers \(0\) to \(\pi\). This constraint ensures that the functions are true inverses.

For differentiation, inverse functions transform expressed derivatives into ones that account for the angle's derivative with respect to the variable. This integration of inverse functions into calculus establishes foundational understanding, critical in advanced topics like solving equations or optimizing functions.