Problem 34 - Calculus

Question

Compute \(d y / d x\) by differentiating \(\ln y .\) This is LD: $$ y=(\sqrt{x})(\sqrt[3]{x})(\sqrt[6]{x}) $$

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} \) is 1.

Step 1: Express the Function in Exponential Form

Rewrite the given function \( y = \sqrt{x} \sqrt[3]{x} \sqrt[6]{x} \) in terms of exponents to simplify differentiation. We have \( y = x^{1/2} \times x^{1/3} \times x^{1/6} \).

Step 2: Combine the Exponents

Combine the exponents using the property \( a^m \times a^n = a^{m+n} \). Add the exponents: \( \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1 \). Thus, the equation simplifies to \( y = x^1 \).

Step 3: Apply the Natural Logarithm

Take the natural logarithm of both sides to facilitate differentiation. We have \( \ln y = \ln(x^1) \).

Step 4: Differentiate Both Sides with Respect to x

Differentiate the equation \( \ln y = \ln x \) with respect to \( x \). The derivative of \( \ln y \) with respect to \( x \) is \( \frac{1}{y} \cdot \frac{dy}{dx} \), and the derivative of \( \ln x \) with respect to \( x \) is \( \frac{1}{x} \).

Step 5: Solve for \( \frac{dy}{dx} \)

Equate the derivatives from the previous step to obtain \( \frac{1}{y} \cdot \frac{dy}{dx} = \frac{1}{x} \). Solve for \( \frac{dy}{dx} \) to get \( \frac{dy}{dx} = \frac{y}{x} \).

Step 6: Substitute Back the Original Expression for y

Since \( y = x^1 = x \), substitute back into \( \frac{dy}{dx} = \frac{y}{x} \) to obtain \( \frac{dy}{dx} = \frac{x}{x} = 1 \).

Key Concepts

Understanding Natural Logarithm in Differentiation

The natural logarithm, denoted by \( \ln \), is a mathematical function that serves as the inverse of the exponential function with base \( e \), where \( e \) is approximately equal to 2.718. In calculus, the natural logarithm is often used to simplify expressions before finding their derivatives, especially when dealing with products, quotients, or powers.
When differentiating a function like \( \ln y \), it helps to convert multiplication of variables into addition of logs, using the identity \( \ln(a \times b) = \ln a + \ln b \). This property is particularly useful when you're handling complex expressions that involve multiplication or division of variables in exponential form.
In our exercise, taking the natural log of both sides is a strategy to simplify differentiation of the product \( \sqrt{x} \sqrt[3]{x} \sqrt[6]{x} \) by transforming it into a single term expression of \( \ln y = \ln(x^1) \), which is easier to work with in calculus.

Working with Exponents

Exponents are a way of expressing repeated multiplication of a number by itself. When you see a term like \( x^{1/2} \), it represents the square root of \( x \), while \( x^{1/3} \) is the cube root.
Exponents follow certain rules which make it easier to handle complex expressions. For example, one key property is that when you multiply similar bases, you can add their exponents: \( a^m \times a^n = a^{m+n} \).
In the problem at hand, we had to simplify \( \sqrt{x} \sqrt[3]{x} \sqrt[6]{x} \) by rewriting it as \( x^{1/2} \times x^{1/3} \times x^{1/6} \) and then adding the exponents together to form \( x^1 \). This simplification paves the way for straightforward differentiation.

The Derivative Explained

A derivative represents the rate at which a function changes as its input changes; essentially, it measures the 'slope' of a function. When you differentiate \( \ln y \) with respect to \( x \), you are finding how \( y \) changes as \( x \) changes.
According to the chain rule, the derivative of \( \ln y \) with respect to \( x \) is \( \frac{1}{y} \cdot \frac{dy}{dx} \). This occurs because \( y \) is a function of \( x \), meaning differentiating \( \ln y \) involves understanding how changes in \( x \) affect \( y \).
This concept is useful in the exercise as it helps in equating and finding the derivative \( \frac{dy}{dx} \), thus solving for how our original expression of \( y \) changes concerning \( x \).

Exponential Form in Differentiation

Exponential form is a mathematical notation that expresses numbers as a base raised to an exponent. Converting a function into exponential form aids in unraveling compound functions for differentiation.
In this exercise, converting \( y \) to an exponential form helps in unraveling the complexity of taken roots which eases the differentiation process. Here, \( \sqrt{x} \sqrt[3]{x} \sqrt[6]{x} \) is converted to \( x^{1/2} \times x^{1/3} \times x^{1/6} \), simplifying it to \( x^1 \).
Thus, transforming complex expressions to a simple exponential form proves beneficial for easier derivation, enabling us to apply basic differentiation techniques effectively.