Problem 55 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=(t \tan t)^{10}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dt} = 10 (t \tan t)^9 (\tan t + t \sec^2 t) \)."

Step 1: Understand the Function

The given function is \( y = (t \tan t)^{10} \). We need to differentiate it with respect to \( t \) to find \( \frac{dy}{dt} \). This function is a power of a product of \( t \) and \( \tan t \).

Step 2: Apply the Chain Rule

To differentiate \( y = (t \tan t)^{10} \), use the chain rule: \( \frac{d}{dt}[u^n] = n \, u^{n-1} \frac{du}{dt} \), where \( u = t \tan t \) and \( n = 10 \). First, find \( \frac{du}{dt} \).

Step 3: Differentiate the Inner Function

Find \( \frac{du}{dt} \) where \( u = t \tan t \). Use the product rule here, which is \( \frac{d}{dt}[fg] = f' g + f g' \). Here, \( f(t) = t \) and \( g(t) = \tan t \), thus \( f' = 1 \) and \( g' = \sec^2 t \).

Step 4: Product Rule Application

Using the product rule: \( \frac{du}{dt} = 1 \cdot \tan t + t \cdot \sec^2 t = \tan t + t \sec^2 t \).

Step 5: Differentiate Using Chain Rule

Now differentiate \( y = (t \tan t)^{10} \). Applying the chain rule, get \( \frac{dy}{dt} = 10(t \tan t)^9 \cdot \frac{d}{dt}(t \tan t) = 10(t \tan t)^9 (\tan t + t \sec^2 t) \).

Step 6: Simplify the Expression

Distribute terms to simplify. \( \frac{dy}{dt} = 10 (t \tan t)^9 (\tan t) + 10(t \tan t)^9 (t \sec^2 t) \).

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus differentiation. It is used when you want to differentiate a composite function. This is a function where one function is nested inside another. Consider a function of the form \(y = [f(g(t))]^n\). The chain rule tells us how to differentiate this by multiplying the derivative of the outer function by the derivative of the inner function.
For example, in the original exercise, we needed to differentiate \(y = (t \tan t)^{10}\). Here, the outer function is \(u^{10}\) and the inner function is \(u = t \tan t\). So, the derivative according to the chain rule is:

• Differentiate the outer function: \(10u^{9}\).
• Find the derivative of the inner function \(\frac{du}{dt}\).

Multiply the derivative of the outer function by \(\frac{du}{dt}\) to get the final result. Thus, the chain rule simplifies the process of dealing with composite functions by breaking it down into manageable parts.

Product Rule

The product rule is another rule in calculus differentiation used when you have the product of two functions that need to be differentiated. It states that if you have two functions \(f(t)\) and \(g(t)\), then the derivative of their product is given by:
\[ \frac{d}{dt}[f(t)g(t)] = f'(t)g(t) + f(t)g'(t) \]
This means you take the derivative of the first function and multiply it by the second function, and then add the product of the first function and the derivative of the second function.
In the original exercise, the inner function \(u = t \tan t\) required the application of the product rule:

• Let \(f(t) = t\) and \(g(t) = \tan t\).
• Differentiate \(t\) to get \(1\).
• Differentiate \(\tan t\) to get \(\sec^2 t\).
• Apply the product rule to get \(\frac{du}{dt} = 1 \cdot \tan t + t \cdot \sec^2 t\).

Using these calculations, you can efficiently manage and simplify complex derivatives by focusing on one piece at a time.

Trigonometric Functions

Trigonometric functions like \(\tan t\) and \(\sec t\) are fundamental in calculus differentiation. Knowing their derivatives is crucial for solving problems involving trigonometric expressions. The derivatives of the basic trigonometric functions are:

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

In the exercise, recognizing that \(g(t) = \tan t\) and its derivative being \(\sec^2 t\) allowed us to apply the product rule effectively. Understanding these derivatives is not only useful for differentiation, it also helps in solving integrals and other calculus problems.
By familiarizing yourself with these rules and derivatives, you can approach calculus problems with greater confidence.