Question

In Exercises \(25-28\) find \(d r / d \theta\). $$r=\sec 2 \theta \tan 2 \theta$$

Short answer

The derivative is \( \frac{{dr}}{{d\theta}} = 2\sec^3 2\theta + 2\sec 2\theta \tan^2 2\theta \)

Step-by-step solution

- Identifying the functions

Identify the functions to be differentiated using the product rule. Here, \( u = \sec 2\theta \) and \( v = \tan 2\theta \) are the functions.

- Differentiate using the product rule

Apply the product rule for differentiation, which states that the derivative of two multiplied functions is the first function times the derivative of the second function plus the second function times the derivative of the first function. Using this rule, \( \frac{{dr}}{{d\theta}} = u \frac{{dv}}{{d\theta}} + v \frac{{du}}{{d\theta}} \).

- Differentiate using the chain rule

Differentiate \( u \) and \( v \) with respect to \( \theta \) using the chain rule. The chain rule states that the derivative of a composite function is the derivative of the outside function times the derivative of the inside function.

- Compute the derivatives

Compute the derivatives to obtain \( \frac{{du}}{{d\theta}} = 2\sec 2\theta \tan 2\theta \) and \( \frac{{dv}}{{d\theta}} = 2\sec^2 2\theta \). Substituting these values into the equation obtained from Step 2, we get \( \frac{{dr}}{{d\theta}} = \sec 2\theta * 2\sec^2 2\theta + \tan 2\theta * 2\sec 2\theta \tan 2\theta \).

- Simplify the result

Simplify to obtain the final result, \( \frac{{dr}}{{d\theta}} = 2\sec^3 2\theta + 2\sec 2\theta \tan^2 2\theta \)

Key concepts

Chain Rule

The chain rule is a fundamental principle in calculus used for finding the derivative of composite functions. A composite function is formed when one function is applied to the result of another, such as in the given exercise where we have trigonometric functions multiplied by each other that are functions of a doubled angle. When using the chain rule, you differentiate the outer function as if the inner function were just a variable, and then multiply that result by the derivative of the inner function.

For example, if we have a function of the form \(f(g(x))\), its derivative, as dictated by the chain rule, is \(f'(g(x))g'(x)\). In our exercise, to differentiate \(u = \(sec 2\theta\)) and \(v = tan 2\theta\) with respect to \(\theta\), we treat \(2\theta\) as the inner function. Therefore, the derivative of \(u\), \(du/d\theta\), involves differentiating \(\sec(2\theta)\) with respect to \(2\theta\), then multiplying by the derivative of \(2\theta\) with respect to \{\theta\), which is 2. This yields \(2\sec 2\theta \tan 2\theta\).

Trigonometric Differentiation

Trigonometric differentiation deals with finding the derivative of trigonometric functions, such as sine, cosine, tangent, and their reciprocals, secant, cosecant, and cotangent. Each of these functions has its own specific rule for differentiation. For instance, the derivatives of sine and cosine are cosine and negative sine, respectively.

To differentiate more complex trigonometric functions, you often need to apply multiple rules. In the exercise, we differentiate the secant and tangent functions. The derivative of \(\sec(x)\) is \(\sec(x)\tan(x)\), and the derivative of \(\tan(x)\) is \(\sec^2(x)\). We apply these rules after using the chain rule to account for the inner function \(2\theta\), as mentioned previously. This is crucial because trigonometric functions often come in the form of composite functions in calculus, making the chain rule a commonly used companion to trigonometric differentiation.

Implicit Differentiation

Understanding Implicit Differentiation

Implicit differentiation is used when dealing with equations where the dependent variable cannot be easily solved for and expressed explicitly as a function of the independent variable. In such cases, we differentiate both sides of the equation with respect to the independent variable while applying the chain rule to any term involving the dependent variable.

This technique comes in handy when dealing with complex equations that define y implicitly, rather than explicitly as \(y=f(x)\). In some problems, trigonometric identities can convert an implicit function into an explicit one, after which we can differentiate directly. However, in our exercise, explicit differentiation is straightforward as we have an explicit function in terms of \(\theta\). Nevertheless, understanding implicit differentiation broadens the ability to tackle a wider range of problems, including those where re-arranging the equation to make \(y\) the subject isn't possible or practical.