Problem 36 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ r=\sec \sqrt{\theta} \tan \left(\frac{1}{\theta}\right) $$

Step-by-Step Solution

Answer

The derivative is \(r' = \frac{1}{2\sqrt{\theta}} \sec\sqrt{\theta} \tan\sqrt{\theta} \tan\left(\frac{1}{\theta}\right) - \frac{\sec^3\left(\frac{1}{\theta}\right)}{\theta^2} \sec\sqrt{\theta}\).

Step 1: Identify the Rule of Differentiation Needed

To differentiate the function \(r = \sec\sqrt{\theta} \tan\left(\frac{1}{\theta}\right)\), we need to use the product rule since the function is a product of two separate functions: \(\sec\sqrt{\theta}\) and \(\tan\left(\frac{1}{\theta}\right)\). Additionally, within these, we need the chain rule for better precision.

Step 2: Differentiate \(\sec\sqrt{\theta}\)

Start by setting \(u = \sec\sqrt{\theta}\). The chain rule applies: it requires differentiating \(\sec v\) where \(v = \sqrt{\theta}\). The derivative of \(\sec v\) is \(\sec v \tan v\). Also, \(v' = \frac{1}{2\sqrt{\theta}}\). Combining, the derivative is: \[\frac{d}{d\theta}(\sec\sqrt{\theta}) = \sec\sqrt{\theta} \tan\sqrt{\theta} \cdot \frac{1}{2\sqrt{\theta}}\].

Step 3: Differentiate \(\tan\left(\frac{1}{\theta}\right)\)

Setting \(v = \tan\left(\frac{1}{\theta}\right)\), apply the chain rule: the derivative of \(\tan u\) is \(\sec^2 u\), and the derivative of \(\frac{1}{\theta}\) is \(-\frac{1}{\theta^2}\). Thus, the derivative is: \[\frac{d}{d\theta}(\tan(\frac{1}{\theta})) = \sec^2\left(\frac{1}{\theta}\right)\cdot \left(-\frac{1}{\theta^2}\right)\].

Step 4: Apply the Product Rule

Using the product rule \((uv)' = u'v + uv'\), combine the derivatives found in steps 2 and 3. For \(u = \sec\sqrt{\theta}\) and \(v = \tan\left(\frac{1}{\theta}\right)\), substitute:\[\begin{align*}r' &= (\sec\sqrt{\theta} \tan\sqrt{\theta} \cdot \frac{1}{2\sqrt{\theta}}) \cdot \tan\left(\frac{1}{\theta}\right) \&+ \sec\sqrt{\theta} \cdot \left(\sec^2\left(\frac{1}{\theta}\right)\cdot \left(-\frac{1}{\theta^2}\right)\right).\end{align*}\]

Step 5: Simplify the Expression

Combine and simplify the terms:\[r' = \frac{1}{2\sqrt{\theta}} \sec\sqrt{\theta} \tan\sqrt{\theta} \tan\left(\frac{1}{\theta}\right) - \frac{1}{\theta^2} \sec\sqrt{\theta} \sec^2\left(\frac{1}{\theta}\right)\]

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used to find the derivative of a product of two functions. When you have a function that is the product of two functions, say \(f(x)\) and \(g(x)\), the product rule states that the derivative of this product is not simply the product of the derivatives of each function. Instead, it is given by the formula: \[(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\] This means you take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function.

• Useful when dealing with expressions like \(r = \sec\sqrt{\theta} \tan(\frac{1}{\theta})\).
• Requires knowledge of both functions involved for accurate results.

In our original problem, we apply the product rule to \(\sec\sqrt{\theta}\) and \(\tan(\frac{1}{\theta})\). By doing so, we're able to decompose a seemingly complex function into manageable parts, each of which can individually be differentiated before applying the rule.

Chain Rule

The chain rule is another essential technique in calculus, especially useful when differentiating composite functions. A composite function is a function applied inside another function, denoted as \(f(g(x))\). The chain rule allows us to take the derivative of these by combining both their derivatives in a sequence. The chain rule formula is summarized as: \[(f(g(x)))' = f'(g(x)) \, g'(x)\] This means you first differentiate the outer function, evaluated at the inner function, and then multiply by the derivative of the inner function.

• Used in differentiating nested functions like \(\tan(\frac{1}{\theta})\) and \(\sec\sqrt{\theta}\).
• Essential for accuracy in expressions with layers of operations.

In our derivative problem, the chain rule helps differentiate \(\sec\sqrt{\theta}\) by initially treating \(\sqrt{\theta}\) as a separate entity. Similarly, it handles the inner function \(\frac{1}{\theta}\) within \(\tan\left(\frac{1}{\theta}\right)\). Mastering the chain rule can simplify the process of working with any function that stacks multiple layers on one another.

Trigonometric Functions

Trigonometric functions play a crucial role in many calculus problems, especially when differentiation is involved. Understanding their derivatives is fundamental in calculus. The most commonly used trigonometric functions include \(\sin(x)\), \(\cos(x)\), and \(\tan(x)\), each with properties like periodicity and specific derivatives:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).
• The derivative of \(\tan(x)\) is \(\sec^2(x)\).

Understanding these derivatives allows you to differentiate trigonometric expressions accurately and efficiently. In the original exercise, the functions \(\sec(x)\) and \(\tan(x)\) were centered to differentiate. The derivative of \(\sec(x)\) is \(\sec(x)\tan(x)\), while as mentioned before the derivative of \(\tan(x)\) is \(\sec^2(x)\). These derivatives directly apply when finding the derivative of expressions like \(r = \sec\sqrt{\theta} \tan\left(\frac{1}{\theta}\right)\). Hence, a strong grip on these principles can drastically simplify differentiation tasks involving trigonometric components.