Problem 11 - Calculus Early Transcendentals

Question

Use Theorem 7 to find the following derivatives. When feasible, express your answer in terms of the independent variable. $$d w / d t, \text { where } w=\cos 2 x \sin 3 y, x=t / 2, \text { and } y=t^{4}$$

Step-by-Step Solution

Answer

Question: Find the derivative of \(w = \cos(2x)\sin(3y)\) with respect to \(t\), given that \(x = \frac{t}{2}\) and \(y = t^4\). Answer: \(\frac{dw}{dt} = -\sin(t)\sin(3t^4) + 12t^3\cos(t)\cos(3t^4)\)

Step 1: Differentiate w with respect to x and y

We have \(w = \cos(2x)\sin(3y)\). Using the product rule and chain rule, we get: $$\frac{dw}{dx} = -2\sin(2x)\sin(3y)$$ $$\frac{dw}{dy} = 3\cos(2x)\cos(3y)$$

Step 2: Differentiate x and y with respect to t

Given \(x = \frac{t}{2}\) and \(y = t^4\), we differentiate these functions with respect to \(t\): $$\frac{dx}{dt} = \frac{1}{2}$$ $$\frac{dy}{dt} = 4t^3$$

Step 3: Apply the chain rule

Now we will apply the chain rule formula and plug in the derivatives obtained in Steps 1 and 2: $$\frac{dw}{dt} = \frac{dw}{dx} \cdot \frac{dx}{dt} + \frac{dw}{dy} \cdot \frac{dy}{dt}$$ $$\frac{dw}{dt} = (-2\sin(2x)\sin(3y)) \cdot \frac{1}{2} + (3\cos(2x)\cos(3y)) \cdot 4t^3$$

Step 4: Simplify and express the answer in terms of t

Replace \(x\) and \(y\) with the expressions in terms of \(t\), and simplify: $$\frac{dw}{dt} = (-2\sin(2(\frac{t}{2}))\sin(3(t^4))) \cdot \frac{1}{2} + (3\cos(2(\frac{t}{2}))\cos(3(t^4))) \cdot 4t^3$$ $$\frac{dw}{dt} = (-\sin(t)\sin(3t^4)) + 12t^3\cos(t)\cos(3t^4)$$ The derivative of \(w\) with respect to \(t\) is: $$\frac{dw}{dt} = -\sin(t)\sin(3t^4) + 12t^3\cos(t)\cos(3t^4)$$

Key Concepts

Product Rule

The Product Rule is a fundamental concept in calculus used when differentiating functions that are products of two other functions. Suppose you have a function \( w(x, y) = f(x)g(y) \), where both \( f \) and \( g \) are differentiable functions. To find the derivative \( \frac{d}{dx}[f(x)g(x)] \), the Product Rule states:

• Take the derivative of the first function and multiply it by the second function.
• Add that to the product of the first function and the derivative of the second function.

In formula terms, this is expressed as:\[ \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) \]
In our original exercise, we are applying the Product Rule to \( w = \cos(2x)\sin(3y) \). Here, \( \cos(2x) \) can be seen as one function and \( \sin(3y) \) as another. By the product rule:

• First, differentiate \( \cos(2x) \) with respect to \( x \), resulting in \( -2\sin(2x) \).
• Second, the \( \sin(3y) \) remains as it is.
• Plus, express \( \cos(2x) \) and differentiate \( \sin(3y) \) with respect to \( y \), which yields \( 3\cos(3y) \).

Thus, you can break a complex multiplication down into manageable parts by using the Product Rule.

Implicit Differentiation

Implicit differentiation is a technique used when it is difficult or impossible to solve for one variable explicitly in terms of another. Sometimes, functions are given in forms such as \( F(x, y) = 0 \), where \( y \) is a function of \( x \) but is not isolated. In such cases, you take derivatives on both sides of the equation with respect to \( x \), treating \( y \) as an implicit function of \( x \).

• Whenever you differentiate a term with \( y \), apply the chain rule to ensure the inclusion of \( \frac{dy}{dx} \).
• This concept is akin to differentiating parametric equations, where each variable is a function of another variable, such as \( t \).

Even though we didn't use full implicit differentiation in the original task, the concept helps differentiate \( w \) with respect to two parameters, \( x \) and \( y \). Both are inherently linked through \( t \), requiring us to apply the Chain Rule. This approach is especially useful when dealing with complex equations where direct differentiation isn't straightforward.

Trigonometric Functions

Trigonometric functions such as \( \sin \), \( \cos \), and \( \tan \) often appear in differentiation exercises. Understanding their derivatives is vital for solving calculus problems:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• These functions require familiarity as they frequently appear in both standalone and composite functions.

In the exercise, both \( \cos(2x) \) and \( \sin(3y) \) involve trigonometric functions, each modified by a coefficient inside the angle. The Chain Rule helps to compute these derivatives:

• For \( \cos(2x) \), multiply the derivative of \( \cos \) by the derivative of \( 2x \), giving \( -2\sin(2x) \).
• Similarly, \( \sin(3y) \) becomes \( 3\cos(3y) \) after applying the chain rule to \( 3y \).

A strong grasp of these principles is essential for accurately finding the derivatives in trigonometric-laden expressions.