Problem 4 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Differentiate the following. (a) \(y=\cos ^{2} x\) (b) \(y=\cos \left(x^{2}\right)\) (c) \(y=x \tan ^{2} x\) (d) \(y=\sin ^{3}\left(x^{4}\right)\) (e) \(y=7[\cos (5 x)+3]^{x}\)

Step-by-Step Solution

Answer

The derivatives of the given functions are as follows:(a) \(-2\cos x*\sin x\)(b) \(-2x*\sin (x^{2})\)(c) \(1*\tan^{2} x + x *2\tan x * \sec ^{2} x\)(d) \(12x^{3}* \sin^{2}(x^{4})*\cos (x^{4})\)(e) \(7*x*[\cos(5x)+3]^{(x-1)}*(1+\ln[\cos(5x)+3]*[\cos(5x)+3])\)}

Step 1: Differentiate \(y=\cos ^{2} x\)

Use the chain rule and the derivative of cosine. Chain rule is applied if a function is composed of multiple functions. Derivative of \(\cos x\) is \(-\sin x\) and when we use chain rule, the expression becomes \(-2\cos x*\sin x\).

Step 2: Differentiate \(y=\cos \left(x^{2}\right)\)

Apply the chain rule again due to the composition of functions. The derivative of \(x^2\) is \(2x\), thus resulting in the expression \(-2x*\sin (x^{2})\).

Step 3: Differentiate \(y=x \tan ^{2} x\)

The product rule is needed here because the function is a product of two differentiable functions. The product rule states that the derivative of u*v with respect to x equals to \(u'v + uv'\). The derivative of \(x\) is \(1\) and of \(\tan ^{2} x\) is \(2\tan x * \sec ^{2} x\), thus resulting in \(1*\tan^{2} x + x *2\tan x * \sec ^{2} x\).

Step 4: Differentiate \(y=\sin ^{3}\left(x^{4}\right)\)

The chain rule is applied several times due to multiple compositions of functions. The derivative of \(\sin^{3} (x^{4})\) is \(3\sin^{2} (x^{4})*\cos (x^{4})\), which when further differentiated using chain rule gives the expression \(3\sin^{2}(x^{4})*\cos (x^{4})*4x^{3}\).

Step 5: Differentiate \(y=7[\cos (5 x)+3]^{x}\)

This requires use of both the chain rule and the power rule, with the understanding that the derivative of \(\cos (5x)\) is \(-5*\sin (5x)\). The power rule for functions in the form of \(a^{u}\), where a is a constant and u is a function of x, is given by \(ln(a)*a^{u}*u'\), thus simplifying the expression to \(7*x*[\cos(5x)+3]^{(x-1)}*(1+\ln[\cos(5x)+3]*[\cos(5x)+3])\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus when dealing with composite functions. Think of it as a way to "unpack" functions that are nested within each other, similar to peeling layers of an onion. Suppose you have a function y that depends on u, which in turn depends on x. In that case, the derivative of y with respect to x can be found using the chain rule, which states:

• If you have a function of the form y = f(g(x)), then the derivative is given by \[ \frac{dy}{dx} = f'(g(x))\cdot g'(x) \]

Let's expand the context with a practical example: differentials of \( y=\cos^{2} x \) requires use of the chain rule. Here, the external function is \( u^2 \) and the inner function is \( u = \cos x \). To differentiate, first find the derivative of the outer function with respect to u, which is \( 2u \), and then multiply it by the derivative of the inner function, \( -\sin x \), resulting in\( -2 \cos x \cdot \sin x \).
Understanding the chain rule enables you to systematically tackle composite functions by breaking them down into manageable parts and subsequently differentiating each part. It transforms complex differentiation tasks into simpler steps, allowing for step-by-step problem solving.

Product Rule

The product rule is indispensable when differentiating functions that are multiplied together. It simplifies the process of finding the derivative of two differentiable functions. According to the product rule, if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x) \cdot v(x) \) is \[ \frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\].Consider the example \( y = x \tan^{2} x \). Here, we have:

• \( u = x \), with its derivative \( u' = 1 \)
• \( v = \tan^{2}x \), with its derivative \( v' = 2\tan x \cdot \sec^{2} x \)

Using the product rule, combine these derivatives:\( 1 \cdot \tan^{2} x + x \cdot 2 \tan x \cdot \sec^{2} x \).
This process illustrates how differentiating products of functions can be made straightforward and methodical with the product rule. Just like a recipe, it provides clear guidelines for success without overwhelming complexity.

Derivative of Trigonometric Functions

Differentiating trigonometric functions is a cornerstone in calculus, pivotal for understanding broader applications in physics, engineering, and beyond. Trigonometric functions such as \( \sin x \), \( \cos x \), and \( \tan x \) come with their own set of rules that simplify differentiating them. They are:

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

When these functions are nested within other functions, the rules of differentiation such as the chain rule become crucial.
Looking at the example \( y = \sin^{3}(x^{4}) \), we apply the derivative rule for sine first: \( 3\sin^{2}(x^{4}) \cdot \cos(x^{4}) \). Next, apply the chain rule to the inner function \( x^{4} \) which results in \( 4x^{3} \). The final expression becomes:\( 3 \sin^{2}(x^{4})\cos(x^{4})\cdot4x^{3} \).Understanding and mastering the derivatives of trigonometric functions allow for seamless application in more complex scenarios and problems. It's like having a map that guides you through the often twisty roads of math.