Problem 71 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the given independent variable. $$y=x^{3} \log _{10} x$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = 3x^2 \log_{10} x + \frac{x^2}{\ln 10}\).

Step 1: Identify the Function and Variable

We are given the function \(y = x^3 \log_{10} x\) and need to find the derivative with respect to the variable \(x\). We will apply the product rule since the function is the product of two simpler functions: \(x^3\) and \(\log_{10} x\).

Step 2: Apply the Product Rule

The product rule states that the derivative of the product of two functions \(u\) and \(v\) is \(u'v + uv'\). Here, let \(u = x^3\) and \(v = \log_{10} x\). We need to find \(u'\) and \(v'\).

Step 3: Differentiate \(u = x^3\)

The derivative of \(u = x^3\) with respect to \(x\) is \(u' = 3x^2\).

Step 4: Differentiate \(v = \log_{10} x\)

To differentiate \(v = \log_{10} x\), convert it into a natural logarithm: \(\log_{10} x = \frac{\ln x}{\ln 10}\). The derivative, using the chain rule, is \(v' = \frac{1}{x \ln 10}\).

Step 5: Substitute Back into the Product Rule

Using the results from Steps 3 and 4, substitute into the product rule formula: \[ \frac{dy}{dx} = u'v + uv' = (3x^2)(\log_{10} x) + (x^3)\left(\frac{1}{x \ln 10}\right) \].

Step 6: Simplify the Expression

Simplify the expression: \[ \frac{dy}{dx} = 3x^2 \log_{10} x + \frac{x^2}{\ln 10} \]. The expression is now fully simplified.

Key Concepts

Product Rule

The product rule is a fundamental concept in differentiation used when we want to differentiate a product of two functions. In this exercise, we are given a function that is a product of two separate functions: \( x^3 \) and \( \log_{10} x \). The product rule helps us to find the derivative of this complete function by first finding the derivatives of the individual parts and then combining them in a specific way.

• The derivative of function \( u \) is denoted as \( u' \).
• The derivative of function \( v \) is denoted as \( v' \).

The product rule formula is:\[(u \cdot v)' = u' \cdot v + u \cdot v'.\]This formula tells us that to find the derivative of \( y = x^3 \log_{10} x \), we compute:

• The derivative of the first function \( x^3 \), which is \( 3x^2 \).
• The derivative of the second function \( \log_{10} x \), which we will calculate using another rule shortly.

We then apply the product rule to combine these results and get the derivative of the entire function.

Chain Rule

The chain rule is essential when we have a function inside another function, which is common during differentiation. In our example, we need the chain rule to find the derivative of \( \log_{10} x \).When you see a function like \( \log_{10} x \), it can be helpful to rewrite this in terms of natural logarithms (ln), because the derivative of \( \ln x \) is straightforward. We express \( \log_{10} x \) as:\[\log_{10} x = \frac{\ln x}{\ln 10}.\]Using the chain rule allows us to differentiate \( \frac{\ln x}{\ln 10} \). Here, the rule helps us recognize that:

• The outer function is a constant multiple, \( \frac{1}{\ln 10} \).
• The inner function is \( \ln x \).

To differentiate:

• The derivative of \( \ln x \) is \( \frac{1}{x} \).

Thus, the derivative of \( \log_{10} x \) becomes:\[v' = \frac{1}{x \ln 10}.\]In the product rule, we need this derivative to complete our computation of the overall derivative.

Logarithmic Differentiation

Logarithmic differentiation becomes particularly useful when differentiating products or quotients of powers and exponential functions. It provides a systematic way to handle complicated functions by reducing them through the power of natural log properties into simpler differentiable terms.In this exercise, while we didn't fully utilize logarithmic differentiation, we were briefly touching on it by converting \( \log_{10} x \) to natural logs. Here is how you typically use logarithmic differentiation:

• Take the natural logarithm of both sides of an equation \( y = f(x) \).
• Use log properties to simplify the expression.
• Differentiate both sides which often results in the product or chain rule application.

With our function \( y = x^3 \log_{10} x \), simplifying \( \log_{10} x \) into a form that is easier to differentiate was just a brief application of these ideas to help us get to the final differentiated expression more efficiently.