Question

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

Short answer

The derivative of \(r\) with respect to \(\theta\), \(\frac{d r}{d \theta}\), is \(- \sec^2(2-\theta)\).

Step-by-step solution

Identify the Outer Function

In the function \(r=\tan (2-\theta)\), the outer function is \(\tan(x)\). The derivative of \(\tan(x)\) with respect to \(x\) is represented as \(1 / \cos^2(x)\).

Identify the Inner Function

In the function \(r=\tan (2-\theta)\), the inner function is \(2-\theta\). The derivative of \(2-\theta\) with respect to \(\theta\) is -1.

Apply the Chain Rule

The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In this case, \(\frac{d r}{d \theta} =\frac{d \tan (2-\theta)}{d \theta} = \frac{d \tan (2-\theta)}{d (2-\theta)} * \frac{d (2-\theta)}{d \theta }\). Substituting the results from Steps 1 and 2 into this formula gives \(\frac{d r}{d \theta} =\frac{1}{ \cos^2 (2-\theta)} * -1 = - \sec^2(2-\theta)\), where \(\sec(x) = \frac{1}{\cos(x)}\).

Key concepts

Derivative of a Composite Function

Understanding the derivative of a composite function is crucial for calculus students. A composite function is, simply put, a function inside another function. If we denote our outer function as \(f\) and the inner function as \(g\), the composite function can be written as \(f(g(x))\).

To find the derivative of a composite function, we use the chain rule, which can be formulated as follows: if \(r = f(g(x))\), then \(\frac{dr}{dx} = f'(g(x)) \times g'(x)\). This means we first calculate the derivative of the outer function with respect to its inner function and then multiply it by the derivative of the inner function with respect to \(x\).

For example, if we have a composite function such as \(r = f(g(x)) = \tan(g(x))\), to find its derivative, we calculate the derivative of \(\tan(x)\)—which is \(\frac{1}{\text{cos}^2(x)}\) or \(\text{sec}^2(x)\)—and multiply it by the derivative of the inner function \(g(x)\). The end result gives us a formula to compute the rate of change of \(r\) with respect to \(x\), which can show the sensitivity of \(r\) to changes in \(x\). This is a fundamental concept in not just calculus, but in all of applied mathematics.

Trigonometric Functions Differentiation

Trigonometric functions such as sine, cosine, and tangent are frequently encountered in calculus problems, and knowing how to differentiate these functions is vital. The derivatives of basic trigonometric functions are standard and must be memorized for efficient problem solving. Here are the key derivatives:

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

When differentiating trigonometric functions, it's important to remember that the chain rule often comes into play, especially when our trigonometric functions involve a composition of functions, like \(\text{tan}(g(x))\). In such cases, we would apply the derivative of \(\text{tan}(x)\) mentioned above and then multiply by the derivative of \(g(x)\). This attention to the components of our function ensures we get accurate results and reinforces our understanding of how trigonometric functions behave under differentiation.

Implicit Differentiation

There are scenarios in calculus where you have to deal with equations where \(y\) is not explicitly expressed as a function of \(x\). In such cases, we use implicit differentiation to find the derivative of \(y\) with respect to \(x\). This technique involves taking the derivative of both sides of an equation with respect to \(x\), while treating \(y\) as a function of \(x\) (even if that function isn't specified).

How to Apply Implicit Differentiation

The process begins by differentiating each term of the equation, applying the chain rule when necessary. If a term includes \(y\), we multiply its derivative by \(\frac{dy}{dx}\). After differentiating, we solve for \(\frac{dy}{dx}\), which may involve algebraically manipulating the equation.

Practical Example

Consider an equation like \(xy = 1\). Differentiating both sides with respect to \(x\) gives us \(x\frac{dy}{dx} + y = 0\). Solving this for \(\frac{dy}{dx}\) would then enable us to find the derivative of \(y\) with respect to \(x\). Implicit differentiation is a powerful tool that expands our capabilities to work with a wider range of functions and relationships in calculus.