Problem 37 - Thomas Calculus

Question

In Exercises \(21-42,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=s \sqrt{1-s^{2}}+\cos ^{-1} s $$

Step-by-Step Solution

Answer

The derivative is \(\frac{s^2}{\sqrt{1-s^2}}\).

Step 1: Differentiate the First Term

The first term in the function is \(s\sqrt{1-s^2}\). We will use the product rule to differentiate this term. The product rule states \((uv)' = u'v + uv'\). Take \(u = s\) and \(v = \sqrt{1-s^2}\). First, differentiate \(u\): \(u' = 1\). Now differentiate \(v\) using the chain rule: \(v = (1-s^2)^{1/2}\) implies \(v' = \frac{1}{2}(1-s^2)^{-1/2}(-2s) = -\frac{s}{\sqrt{1-s^2}}\). Applying the product rule, the derivative of the first term is \(\sqrt{1-s^2} + s\left(-\frac{s}{\sqrt{1-s^2}}\right) = \sqrt{1-s^2} - \frac{s^2}{\sqrt{1-s^2}}\)."

Step 2: Simplify the First Term's Derivative

Combine the terms in the differentiated expression from Step 1: \(\sqrt{1-s^2} - \frac{s^2}{\sqrt{1-s^2}}\). These terms can be combined into a single fraction: \(\frac{1-s^2}{\sqrt{1-s^2}}\)."

Step 3: Differentiate the Second Term

The second term in the function is \(\cos^{-1} s\). The derivative of \(\cos^{-1} s\) with respect to \(s\) is \(-\frac{1}{\sqrt{1-s^2}}\). This comes from the standard derivative rule for the inverse cosine function."

Step 4: Combine the Derivatives

Add the derivatives of the two terms to find the derivative of \(y\): \(\frac{1-s^2}{\sqrt{1-s^2}} - \frac{1}{\sqrt{1-s^2}}\). Simplify this to obtain \(\frac{s^2}{\sqrt{1-s^2}}\)."

Step 5: Final Simplification

Since both terms in the numerator are positive, the expression is already simplified. Therefore, the derivative \(\frac{s^2}{\sqrt{1-s^2}}\) is our final answer."

Key Concepts

Product Rule

The product rule is a fundamental rule in calculus, essential for differentiating products of functions. If you have two functions multiplied together, say \(u(x)\) and \(v(x)\), then the derivative of their product \(uv\) is not simply the product of their derivatives. Instead, the product rule requires a little more effort!

• The product rule can be stated as: \[(uv)' = u'v + uv'\]
• This means you first differentiate \(u\) while keeping \(v\) constant, and then differentiate \(v\) while keeping \(u\) constant. Finally, you add the two results together.

In the given exercise, the function \(y\) included a term \(s \sqrt{1-s^2}\), which necessitated the use of the product rule. Here, \(u = s\) and \(v = \sqrt{1-s^2}\) were chosen, leading to the final derivative term: \(\sqrt{1-s^2} - \frac{s^2}{\sqrt{1-s^2}}\). By using the product rule, you ensure no part of the function is missed in the differentiation process.

Chain Rule

The chain rule is a key tool in calculus for differentiating composite functions. Imagine a function nested inside another function, like \(h(x) = f(g(x))\). The chain rule helps to unravel these layers and find the derivative.

• The chain rule is expressed as: \[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]
• This essentially means you first differentiate the outer function, and then multiply by the derivative of the inner function.

In this particular exercise, the chain rule was critical while differentiating \(v = \sqrt{1-s^2}\). The inner function here is \(1-s^2\) and the outer function is the square root. The derivative was found by first expressing the square root as a power, \((1-s^2)^{1/2}\), then applying the chain rule to get: \[-\frac{s}{\sqrt{1-s^2}}\]The chain rule allows for the breakup of difficult composite functions into simpler parts for differentiation.

Inverse Trigonometric Functions

Inverse trigonometric functions are functions that reverse the action of the standard trigonometric functions. When dealing with derivatives, each inverse trigonometric function has a specific derivative formula that can simplify the differentiation process.

• The inverse cosine function, \(\cos^{-1}(x)\), is relevant in this exercise. Its derivative is given by: \[\frac{d}{dx} [\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}}\]
• This derivative formula comes in handy as it directly gives the rate of change without further manipulation.

In the provided solution, the second component of the function \(y\) was \(\cos^{-1}(s)\). Applying the derivative formula for the inverse cosine, its derivative becomes \(-\frac{1}{\sqrt{1-s^2}}\). This step is crucial in solving the full problem and helps in expressing the complete derivative of the original function \(y\). Understanding how to easily differentiate inverse trigonometric functions simplifies many calculus problems.