Problem 18 - University Calculus: Early Transcendentals

Question

Find \(d y / d x\). $$g(x)=(2-x) \tan ^{2} x$$

Step-by-Step Solution

Answer

The derivative is \( \frac{d}{dx}g(x) = -\tan^2 x + 4\tan x\sec^2 x - 2x\tan x\sec^2 x \).

Step 1: Apply Product Rule

We begin by recognizing that the function \( g(x) = (2-x)\tan^2 x \) is a product of two functions, namely \( u(x) = 2-x \) and \( v(x) = \tan^2 x \). We will apply the product rule for differentiation, which states that if \( g(x) = u(x)v(x) \), then \( \frac{d}{dx}g(x) = u'(x)v(x) + u(x)v'(x) \).

Step 2: Differentiate u(x)

Differentiate \( u(x) = 2-x \) with respect to \( x \). The derivative, \( u'(x) = \frac{d}{dx}(2-x) \), is \( u'(x) = -1 \).

Step 3: Differentiate v(x)

Next, we differentiate \( v(x) = \tan^2 x \) using the chain rule. Recall that \( \frac{d}{dx} [f(x)]^n = n [f(x)]^{n-1} f'(x) \). Let \( f(x) = \tan x \) and \( n = 2 \), so: \( v'(x) = 2 \tan x \cdot \sec^2 x = 2\tan x \cdot (1 + \tan^2 x) = 2\tan x\sec^2 x \).

Step 4: Substitute and Simplify

Substitute \( u(x), u'(x), v(x), \) and \( v'(x) \) into the product rule expression. \( \frac{d}{dx}g(x) = (-1) \cdot \tan^2 x + (2-x) \cdot 2 \tan x \sec^2 x \). Simplify this to: \( \frac{d}{dx}g(x) = -\tan^2 x + 2(2-x)\tan x\sec^2 x \).

Step 5: Final Simplification

We can factor the expression for cosmetic purposes but the core of the answer does not change. The final simplified derivative in standard form is: \( \frac{d}{dx}g(x) = -\tan^2 x + 4\tan x\sec^2 x - 2x\tan x\sec^2 x \).

Key Concepts

Product Rule

The Product Rule is a key differentiation technique used when finding the derivative of the product of two functions. In mathematics, if you have a function defined as the product of two separate functions, say \(u(x)\) and \(v(x)\), then the derivative of this product, according to the Product Rule, is:\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]Here's how it works:

• First, differentiate \(u(x)\) to get \(u'(x)\).
• Then, differentiate \(v(x)\) to find \(v'(x)\).
• Multiply \(u'(x)\) by \(v(x)\), and add to this the product of \(u(x)\) and \(v'(x)\).

In the given exercise, we identify \(u(x) = (2-x)\) and \(v(x) = \tan^2x\). By applying the Product Rule, we calculate the derivative of each function separately and substitute them back into the rule to find the final derivative of the complete expression. This simplifies the process of differentiation of complex products into simpler, manageable steps.

Chain Rule

The Chain Rule is an essential differentiation technique that is invaluable when dealing with compositions of functions. When you have a function within another function, the Chain Rule helps you differentiate the composite function. The rule can be stated as:\[\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)\]This means:

• Differentiate the outer function while keeping the inner function in place.
• Multiply this result by the derivative of the inner function.

In our solution for \(v(x) = \tan^2 x\), we recognize it as a composite function where the outer function is \(f(x) = x^2\) and the inner function is \(g(x) = \tan x\). Thus, using the Chain Rule, we express its derivative as \(2 \tan x \cdot \sec^2 x\). This simplifies a potentially complicated differentiation task into a series of simpler steps by gradually peeling back the layers of nested functions.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, regularly appear in calculus. Differentiating these functions requires understanding their unique derivatives. For example, the derivative of \(\tan x\) is \(\sec^2 x\).Here are a few key tips:

• Always remember that some trigonometric identities can simplify differentiation, such as \(\sec^2 x = 1 + \tan^2 x\).
• Acknowledge that products and powers of trigonometric functions often need both the Product Rule and Chain Rule.

In the given problem, \(\tan^2 x\) was differentiated with the help of identities and rules. Through our step-by-step differentiation, we effectively handle both squaring and multiplication with functions like \(\sec^2 x\) by combining these tools. This combination is especially useful, showing how interconnected these differentiation techniques can be when working with trigonometric expressions.