Question

Evaluating a Second Derivative In Exercises \(91-94\) , evaluate the second derivative of the function at the given point. Use a computer algebra system to verify your result. $$ g(t)=\tan 2 t,\left(\frac{\pi}{6}, \sqrt{3}\right) $$

Short answer

The second derivative of the function at the given point is \(16\sqrt{3}\).

Step-by-step solution

Find the first derivative of the function

First, we need to identify the derivative of \( g(t) = \tan 2t \). Recalling the chain rule, which states that \( \frac{d}{dt} (f(g(t))) = f'(g(t)) \cdot g'(t)\), and that the derivative of \( \tan x \) is \( \sec^2 x \), we can write \[ g'(t) = \sec^2(2t) \cdot 2\].

Find the second derivative of the function

Next, take another derivative to find the second derivative \( g''(t) \) of the function. For any \( x \), consider that \( \frac{d}{dx} \sec^2 x = 2\sec x \cdot \sec x \cdot \tan x \). Using a similar process to Step 1, we find that \[ g''(t) = 4\sec^2(2t) \cdot \tan(2t) \].

Evaluate the second derivative at the given point

Finally, we evaluate our second derivative equation at the point \( t = \frac{\pi}{6} \). Recall the definitions of the secant and tangent functions in terms of sine and cosine, as well as that \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \) and \( \cos \frac{\pi}{3} = \frac{1}{2} \). Substituting these values in, we compute \[ g''\left(\frac{\pi}{6}\right) = 4\left(\frac{1}{\cos^2{\pi}{3}}\right) \cdot \left(\frac{\sin{\pi}{3}}{\cos{\pi}{3}}\right) = 4(4) \cdot \sqrt{3} = 16\sqrt{3}\].

Key concepts

Chain Rule

The chain rule is an essential principle in calculus, used to differentiate composite functions. A composite function is formed when one function is nested inside another, for example, if you have a function like f(g(t)). Think of it as unraveling layers; the chain rule helps you systematically take each layer apart.

When you apply the chain rule, you take the derivative of the outer function, evaluating it at the inner function, and multiply it by the derivative of the inner function itself. The formal rule can be described as:
\[ \frac{d}{dt} f(g(t)) = f'(g(t)) \times g'(t) \].
In our textbook example with g(t) = \tan(2t), the outer function is the tangent function, and the inner function is \(2t\). By using the chain rule, we differentiate \(\tan u\) first (where \(u=2t\)), and then multiply by the derivative of \(2t\), which is simply 2, to find the first derivative of g(t).
Understanding the application of the chain rule is vital, especially when dealing with trigonometric functions that often involve compositions of multiple functions.

Trigonometric Functions

Trigonometric functions are foundational in calculus, particularly in problems involving angles and periodic functions. These functions include sine (\(\sin\)), cosine (\(\cos\)), tangent (\(\tan\)), as well as their reciprocals cosecant (\(\csc\)), secant (\(\sec\)), and cotangent (\(\cot\)). They are essential for modeling oscillatory behavior, as they repeat values in regular intervals and have distinct derivatives.

Each trigonometric function has a specific derivative. For example, the derivative of \(\sin x\) is \(\cos x\) and the derivative of \(\cos x\) is \(-\sin x\). For the tangent function, which we encounter in the exercise, understanding its derivative in terms of secant squared is crucial for solving calculus problems.
Moreover, these functions hold special relationships with one another through their identities, such as the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), which is often used to simplify complex expressions during differentiation and integration.

Derivative of Tan Function

The derivative of the tangent function is particularly interesting and follows a pattern distinct from sine and cosine. If we have \(f(x) = \tan x\), then the derivative of \(f\) with respect to \(x\) is written as \(f'(x)\) or \(\frac{d}{dx}\tan x\).

The derivative of the tangent function is \(\sec^2 x\), represented as:\[ \frac{d}{dx}\tan x = \sec^2 x \].
This result is derived from the quotient rule, as we can express \(\tan x\) as \(\frac{\sin x}{\cos x}\). In our exercise, to find the first derivative of \(\tan(2t)\), we apply the chain rule and obtain \(\sec^2(2t) \times 2\), which simplifies to \(2\sec^2(2t)\).
Mastering this derivative is key, particularly as the tangent and secant functions are often present in problems involving periodic phenomena or angles, and form the backbone of many applied calculus problems.