Problem 52 - Calculus Single Variable

Question

A function \(f\) is given. Use logarithmic differentiation to calculate \(f^{\prime}(x)\). $$ f(x)=\left(x^{2}+1\right)^{3} \sin ^{2}(x) \cos ^{3}(x) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = (x^2 + 1)^3 \sin^2(x) \cos^3(x) \left( \frac{6x}{x^2+1} + 2\cot(x) - 3\tan(x) \right). \)

Step 1: Apply Natural Logarithm

We begin by taking the natural logarithm of both sides of the function to facilitate differentiation using the product and chain rule. Express the function as: \( y = f(x) = (x^2 + 1)^3 \sin^2(x) \cos^3(x) \). Take the natural logarithm: \( \ln(y) = \ln((x^2 + 1)^3) + \ln(\sin^2(x)) + \ln(\cos^3(x)) \) which simplifies using logarithm properties to \( \ln(y) = 3\ln(x^2 + 1) + 2\ln(\sin(x)) + 3\ln(\cos(x)) \).

Step 2: Differentiate Logarithmic Equation

Differentiate both sides with respect to \(x\). On the left side, use implicit differentiation: \( \frac{d}{dx}[\ln(y)] = \frac{1}{y} \cdot \frac{dy}{dx} \). On the right side, differentiate each term: \ \( \frac{d}{dx}[3\ln(x^2+1)] =\frac{6x}{x^2+1} \), \ \( \frac{d}{dx}[2\ln(\sin(x))] = 2 \cdot \frac{\cos(x)}{\sin(x)} = 2 \cot(x) \), \ and \( \frac{d}{dx}[3\ln(\cos(x))] = 3 \cdot (-\frac{\sin(x)}{\cos(x)}) = -3 \tan(x) \)."

Step 3: Combine Results

Combine the results from Step 2 into one derivative equation: \( \frac{1}{y} \cdot \frac{dy}{dx} = \frac{6x}{x^2+1} + 2\cot(x) - 3\tan(x) \).

Step 4: Solve for \(f'(x)\)

To find \(f'(x)\), multiply both sides by \(y\): \( \frac{dy}{dx} = y \left( \frac{6x}{x^2+1} + 2\cot(x) - 3\tan(x) \right) \). Substitute back \(y = f(x) = (x^2 + 1)^3 \sin^2(x) \cos^3(x)\) to get: \[ f'(x) = (x^2 + 1)^3 \sin^2(x) \cos^3(x) \left( \frac{6x}{x^2+1} + 2\cot(x) - 3\tan(x) \right). \]

Key Concepts

Product Rule in Differentiation

Differentiation can sometimes get tricky, especially when dealing with products of functions like in this exercise. This is where the product rule comes into play. The product rule is a technique used to differentiate expressions that are the product of two or more functions. If you have a function that is the product of two functions, say \( u(x) \) and \( v(x) \), the derivative of their product \( y = u(x) \times v(x) \) can be calculated using:

• \( y' = u'(x)v(x) + u(x)v'(x) \)

In our original exercise, \( f(x) = (x^2 + 1)^3 \sin^2(x) \cos^3(x) \), we recognized it as the product of different functions. Instead of directly applying the product rule, we first used logarithmic differentiation, which facilitated breaking the expression into components better handled by logarithmic properties before applying product rule like calculations.

Chain Rule in Calculus

The chain rule is an essential tool in calculus for finding the derivative of compositions of functions. It helps when your function can be split into a composition of two (or more) functions, \( y = g(h(x)) \). To differentiate \( y \) with respect to \( x \), the chain rule suggests:

• \( \frac{dy}{dx} = g'(h(x)) \cdot h'(x) \)

In the context of our exercise, the chain rule is implied when differentiating complex functions like \((x^2 + 1)^3\), where we differentiate the outer function and then multiply by the derivative of the inner function. It’s like peeling back the layers of an onion—differentiating step by step until you reach the "core" function.

Understanding Trigonometric Functions

Trigonometric functions like \( \sin(x) \) and \( \cos(x) \) play a crucial role in calculus, particularly in problems like our exercise. Knowing how these functions behave and differentiate is fundamental. Here are some key derivatives:

• The derivative of \( \sin(x) \) is \( \cos(x) \)
• The derivative of \( \cos(x) \) is \( -\sin(x) \)
• The cotangent function \( \cot(x) = \frac{\cos(x)}{\sin(x)} \) has the derivative \(-\csc^2(x)\)
• The tangent function \( \tan(x) = \frac{\sin(x)}{\cos(x)} \) differentiates to \( \sec^2(x) \)

These derivatives are pivotal for steps that require differentiating parts like \( \sin^2(x) \) and \( \cos^3(x) \) in our original problem, each involving trigonometric identities and transformations.

Implicit Differentiation Made Simple

Implicit differentiation is another technique instrumental in solving differentiation problems, often when equations aren’t solely in terms of one variable. It allows us to find derivatives when functions aren’t explicitly given as \( y = f(x) \). Instead, when you have an equation with \( x \) and \( y \), implicit differentiation facilitates finding \( \frac{dy}{dx} \).Here’s how it works:

• You differentiate every term with respect to \( x \), treating \( y \) as a function of \( x \).
• For derivatives involving \( y \), apply the chain rule, multiply by \( \frac{dy}{dx} \).
• Solve for \( \frac{dy}{dx} \).

In our exercise, implicit differentiation was vital when deriving \( \ln(y) = 3\ln(x^2 + 1) + 2\ln(\sin(x)) + 3\ln(\cos(x)) \), wherein each differentiation step required accounting for the relationships between \( y \) and \( x \). This approach enabled the neat composition of derivatives on the complicated function we analyzed.