Problem 25 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ y=x^{2} \sin ^{4} x+x \cos ^{-2} x $$

Step-by-Step Solution

Answer

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

Step 1: Apply the Product Rule to the First Term

The first term of the function is \(x^2 \sin^4 x\). To find its derivative, use the product rule: \((uv)' = u'v + uv'\) where \(u = x^2\) and \(v = \sin^4 x\). Compute \(u'\) as \(2x\) and \(v'\) using the chain rule.

Step 2: Differentiate Using Chain Rule on Sine Function

For \(v = \sin^4 x\), apply the chain rule. The derivative of \((g(x))^4\) is \(4(g(x))^3 \cdot g'(x)\) where \(g(x) = \sin x\). Thus, the derivative \(v' = 4 \sin^3 x \cos x\).

Step 3: Calculate Derivative of the First Term

Using results from Steps 1 and 2, apply the product rule: \(\frac{d}{dx}(x^2 \sin^4 x) = (2x)(\sin^4 x) + (x^2)(4 \sin^3 x \cos x)\). Simplify to obtain \(2x \sin^4 x + 4x^2 \sin^3 x \cos x\).

Step 4: Differentiate the Second Term

The second term is \(x \cos^{-2} x\). Using the product and chain rules, differentiate \(u = x\), \(u' = 1\), and \(v = \cos^{-2} x\). The derivative \(v' = -2 \cos^{-3} x \cdot (-\sin x) = 2 \sin x \cos^{-3} x\).

Step 5: Calculate Derivative of the Second Term

Apply the product rule: \(\frac{d}{dx}(x \cos^{-2} x) = (1)(\cos^{-2} x) + x(2 \sin x \cos^{-3} x)\). Simplify to get \(\cos^{-2} x + 2x \sin x \cos^{-3} x\).

Step 6: Combine the Derivatives

Add the derivatives from Steps 3 and 5 to find the derivative of the entire function: \(2x \sin^4 x + 4x^2 \sin^3 x \cos x + \cos^{-2} x + 2x \sin x \cos^{-3} x\).

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus for finding the derivative of a product of two functions. Let's break this down to make it clearer for you.
When you have two functions, say \(u(x)\) and \(v(x)\), and you need to differentiate their product \(uv\), the product rule states that:

• The derivative \((uv)'\) is \(u'v + uv'\).

This basically means you differentiate the first function \(u\) and multiply it by the second function \(v\), then add it to the first function \(u\) times the derivative of the second function \(v\).
This rule is especially handy when dealing with polynomials multiplied by other functions, as is often the case in problems involving trigonometric functions or exponential terms. In our exercise, we used the product rule to handle terms like \(x^2 \sin^4 x\) and \(x \cos^{-2} x\).
Remember, apply the product rule separately to each term, and write down every step to avoid confusion!

Chain Rule

The chain rule is a way to differentiate compositions of functions. It’s a bit like peeling back the skin of an onion—layer by layer. You apply this rule when you have a function inside another function, often phrased as 'a function of a function'.
In mathematical terms, if you have a function \(y = f(g(x))\), the chain rule states:

• Its derivative, \(y'\), is \(f'(g(x)) \cdot g'(x)\).

This means you first compute the derivative of the outer function \(f\) at the point \(g(x)\), and then multiply it by the derivative of the inner function \(g\).
In the original problem, we see this with \(\sin^4 x\). Here, the outer function is \(x^4\) and the inner function is \(\sin x\). By applying the chain rule, we get \(4 \sin^3 x \cos x\) for the derivative, as the inner function’s derivative, \(\cos x\), multiplies the derivative of the power rule applied to \(\sin^4 x\).
Practice using the chain rule with simple functions first, and soon you'll be able to tackle more complex compositions seamlessly.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, frequently show up in calculus problems. Knowing how to differentiate these is essential.
For the sine function, \(\sin x\), the derivative is \(\cos x\).
Meanwhile, for the cosine function, \(\cos x\), the derivative is \(-\sin x\). These derivatives are essential when using rules like the product rule or chain rule.
Additionally, when you have functions like \(\sin^n x\) or \(\cos^{-n} x\), you will frequently use the chain rule to differentiate. For example:

• The derivative of \(\sin^4 x\) requires you to think of it as a composition of functions.
• Similarly, for \(\cos^{-2} x\), you will need to apply the chain rule to correctly handle the negative exponent.

In our discussed problem, understanding these basic derivatives and knowing how to apply them will save time and effort, leading to the correct solution efficiently.