Problem 61 - Calculus and its applications

Question

Differentiate. $$ y=x^{6} \ln x-\frac{1}{4} x^{4} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = 6x^5 \ln x + x^5 - x^3 \).

Step 1: Identify the Function Components

The function given to us is \( y = x^6 \ln x - \frac{1}{4}x^4 \). There are two separate terms in this function: the first term is \( x^6 \ln x \) which involves a product of functions and the second term \( -\frac{1}{4}x^4 \) which is a simple polynomial.

Step 2: Differentiate the First Term Using the Product Rule

For the first term \( u = x^6 \) and \( v = \ln x \), use the product rule \( (uv)' = u'v + uv' \). Differentiate each part: \( u' = 6x^5 \) and \( v' = \frac{1}{x} \). Thus, the derivative is \( (x^6 \ln x)' = 6x^5 \ln x + x^6 \cdot \frac{1}{x} = 6x^5 \ln x + x^5 \).

Step 3: Differentiate the Second Term

The second term is \( -\frac{1}{4}x^4 \). Differentiate it using the power rule: \( \frac{d}{dx}[-\frac{1}{4}x^4] = -\frac{1}{4} \cdot 4x^3 = -x^3 \).

Step 4: Combine the Derivatives

Now combine the derivatives from both terms: \[ \frac{dy}{dx} = 6x^5 \ln x + x^5 - x^3 \].

Key Concepts

Product Rule

The product rule is a crucial concept used when differentiating functions that are products of two separate functions. Imagine you have a function that's a product of two parts, like in the exercise, where we have \(u = x^6\) and \(v = \ln x\). When you're asked to differentiate such a function, the product rule tells you the derivative is found as:

• The derivative of the first function \(u\), multiplied by the second function \(v\).
• Then, the first function \(u\) multiplied by the derivative of the second function \(v\).
• Finally, you sum these two results to get \((uv)' = u'v + uv'\).

In our example, differentiate \(u = x^6\) to get \(u' = 6x^5\). Differentiate \(v = \ln x\) to get \(v' = \frac{1}{x}\). Applying the product rule gives \((x^6 \ln x)' = 6x^5 \ln x + x^5\). This process can seem tricky at first, but practicing with different functions will make it second nature.

Power Rule

The power rule is one of the most straightforward techniques for differentiation, especially for polynomial functions. It states that when you have a function of the form \(f(x) = x^n\), its derivative is found by multiplying the power \(n\) by \(x\) and then reducing the power by one:

• Simply apply \( \frac{d}{dx}[x^n] = nx^{n-1} \).

In our problem, for the term \(-\frac{1}{4}x^4\), using the power rule means multiplying the power of 4 by \(-\frac{1}{4}\), and then reducing the exponent by 1. The derivative thus simplifies to \(-x^3\). The clarity of the power rule makes it invaluable for quickly dealing with polynomial expressions.

Polynomial Function

Polynomial functions are expressions made up of terms like \(ax^n\) where \(a\) is a coefficient and \(n\) is a non-negative integer exponent. They are among the most common functions you'll encounter in calculus. A key aspect of polynomial functions is their smooth, continuous nature, allowing them to be easily differentiated using rules like the power rule.Consider the polynomial portion in our exercise: \(-\frac{1}{4}x^4\). This term is already in its simplest polynomial form, and applying differentiation techniques to each part separately helps simplify the process. With polynomial functions, each term \(ax^n\) can independently use the power rule for its derivative.Polynomials are not only easy to differentiate but also serve as building blocks for more complex functions. Mastering them makes handling more complicated functions much more manageable.