Problem 29 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$y=x^{2} \sin ^{4} x+x \cos ^{-2} x$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = 2x \sin^4 x + 4x^2 \sin^3 x \cos x + \cos^{-2} x + \frac{2x \sin x}{\cos^3 x} \).

Step 1: Differentiate each term separately

To find the derivative of the function \(y = x^2 \sin^4 x + x \cos^{-2} x\), we'll apply the derivative rules term by term. The first term is \(x^2 \sin^4 x\) and the second term is \(x \cos^{-2} x\). Use the product rule and chain rule where necessary.

Step 2: Differentiate the first term using the product and chain rule

For the first term \(u = x^2\) and \(v = \sin^4 x\), we use the product rule: \( (uv)' = u'v + uv' \). Here, \(u' = 2x\) and \(v'\) requires the chain rule. For \(v = \sin^4 x\), let \(w = \sin x\) then \(v = w^4\). So \(\frac{dv}{dx} = 4w^3 \cdot \cos x\). Hence, \( \frac{d}{dx} (x^2 \sin^4 x) = 2x \sin^4 x + x^2 \cdot 4 \sin^3 x \cos x \).

Step 3: Differentiate the second term using product and chain rule

For the second term \(w = x\) and \(z = \cos^{-2} x\), apply the product rule again: \( (wz)' = w'z + wz' \). Here, \(w' = 1\) and for \(z = \cos^{-2} x\), we use the chain rule. \(z\) can be rewritten as \((\cos x)^{-2}\), giving \(z' = -2 (\cos x)^{-3} \cdot (-\sin x) = \frac{2 \sin x}{\cos^3 x}\). Thus, \( \frac{d}{dx} (x \cos^{-2} x) = \cos^{-2} x + x \cdot \frac{2 \sin x}{\cos^3 x}\).

Step 4: Combine the derivatives

Combine the derivatives from both terms to get the overall derivative of \(y\). The derivative of the whole function is: \[\frac{dy}{dx} = 2x \sin^4 x + 4x^2 \sin^3 x \cos x + \cos^{-2} x + \frac{2x \sin x}{\cos^3 x}\].

Key Concepts

Product Rule Differentiation

In calculus, the Product Rule is a fundamental method for finding the derivative of a product of two functions. If you have a function composed of two parts, say \( u(x) \) and \( v(x) \), and you need to differentiate \( y = u(x)v(x) \), you use the Product Rule:

• \( (uv)' = u'v + uv' \)

This rule tells us that the derivative of a product is not just the product of the derivatives. Instead, you need to differentiate each function separately and combine them appropriately.
In our exercise, for the function \( x^2 \sin^4 x \), we identify \( u = x^2 \) and \( v = \sin^4 x \). We calculate \( u' = 2x \) and use further rules to find \( v' \), before plugging them into the product rule formula.
This process is crucial in our example, as it helps us deal with compound expressions, ensuring all parts of a product are considered.

Chain Rule Differentiation

The Chain Rule is another key technique used in calculus for differentiating composite functions. When a function is nested within another, such as \( y = (f(g(x))) \), the chain rule is applied:

• \( \frac{dy}{dx} = \frac{dy}{dg} \cdot \frac{dg}{dx} \)

Essentially, it breaks down complex derivatives into simpler ones, by linking the rate of change through derivative connections.
In our problem, to differentiate \( \sin^4 x \), we apply the chain rule. We let \( w = \sin x \), giving \( v = w^4 \). We find \( \frac{dv}{dw} = 4w^3 \) and \( \frac{dw}{dx} = \cos x \), so \( \frac{dv}{dx} = 4 \sin^3 x \cos x \).
This approach helps manage the complexity of nested functions like powers or trigonometric expressions, by focusing on each level of the function separately.

Trigonometric Derivatives

Trigonometric derivatives are essential for functions that involve trigonometric expressions, such as \( \sin x \), \( \cos x \), and more complex ones like \( \tan x \) or \( \sec x \). Each trigonometric function has a standard derivative:

• \( \frac{d}{dx} \sin x = \cos x \)
• \( \frac{d}{dx} \cos x = -\sin x \)
• \( \frac{d}{dx} \tan x = \sec^2 x \)

Understanding these basic derivatives allows us to handle complex trigonometric expressions in a structured manner.
In our exercise, when dealing with \( \cos^{-2} x \), we use the chain rule to find its derivative. Rewriting as \( (\cos x)^{-2} \), we determine the derivative involves both the trigonometric derivative \( \frac{d}{dx} \cos x = -\sin x \) and exponent rules, ultimately yielding \( \frac{2 \sin x}{\cos^3 x} \).
This combination of trigonometric rules ensures that even complex derivatives maintain mathematical accuracy and facilitate understanding in calculus problems.