91Ó°ÊÓ

The product rule is a fundamental tool in calculus for finding the derivative of a product of two functions. It states that if you have two functions, \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x) v(x) \) is \((uv)' = u'v + uv'\). This means that you take the derivative of the first function and multiply by the second, then add the product of the first function and the derivative of the second function.

In the original problem, the function \( r \) is a product of two trigonometric expressions: \( u(\theta) = \sec \sqrt{\theta} \) and \( v(\theta) = \tan \left(\frac{1}{\theta}\right) \).

• First, differentiate \( u(\theta) \) with respect to \( \theta \).
• Then, differentiate \( v(\theta) \) with respect to \( \theta \).
• Finally, apply the product rule to find the derivative of the entire expression.

This approach helps in breaking down complex problems into simpler parts, making it easier to understand and solve.

The chain rule is another crucial concept in calculus used to find the derivative of composite functions. When you have a function inside another function, you'll use the chain rule, which states that if \( y = g(f(x)) \), then the derivative \( y' \) is \( g'(f(x)) \cdot f'(x) \).

In our exercise, the chain rule is applied twice. First, it helps differentiate \( u(\theta) = \sec \sqrt{\theta} \). We substitute \( z = \sqrt{\theta} \) so that \( u(\theta) = \sec(z) \), and differentiate accordingly. The outer function is \( \sec(z) \) and the inner function is \( z = \sqrt{\theta} \).

• The derivative of \( \sec(z) \) is \( \sec(z) \tan(z) \).
• The derivative of \( \sqrt{\theta} \) is \( \frac{1}{2\sqrt{\theta}} \).
• Combine these using the chain rule to find \( u'(\theta) \).

Similarly, for \( v(\theta) = \tan \left(\frac{1}{\theta}\right) \), substitute \( w = \frac{1}{\theta} \). The outer function is \( \tan(w) \) and the inner function is \( w = \frac{1}{\theta} \).

• Differentiate \( \tan(w) \) to get \( \sec^2(w) \).
• The derivative of \( \frac{1}{\theta} \) is \( -\frac{1}{\theta^2} \).
• Use the chain rule for \( v'(\theta) \).

Trigonometric functions like \( \sec \) and \( \tan \) are common in calculus and have well-known derivatives. Knowing these derivatives is essential for handling problems involving trigonometric expressions.

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

In our derivative problem, these derivatives play a key role. For \( u(\theta) = \sec \sqrt{\theta} \), the derivative involves \( \sec \sqrt{\theta} \tan \sqrt{\theta} \). For \( v(\theta) = \tan \left(\frac{1}{\theta}\right) \), the derivative involves \( \sec^2 \left(\frac{1}{\theta}\right) \).

Trigonometric identities and derivatives simplify calculations. They also help in combining terms effectively to obtain the final simplified form for the derivative \( r' \) in the exercise. Understanding and memorizing these derivatives allows you to tackle calculus problems efficiently, ensuring each step is clear and logical.