Question

For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$ y=\sqrt{x} \sqrt[3]{x} \sqrt[6]{x} $$

Short answer

The solution gives \( \frac{dy}{dx} = 1 \).

Step-by-step solution

Write the original function in terms of powers

The given function can be expressed using powers:\[ y = x^{1/2} \cdot x^{1/3} \cdot x^{1/6} \]Combine the exponents: \[ y = x^{1/2 + 1/3 + 1/6} \]

Simplify the exponents

Calculate the sum of the exponents:\[ \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \]Convert the fractions to have a common denominator:\[ \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1 \]Thus, we have \( y = x^1 = x \).

Differentiate by taking natural log

Take the natural logarithm on both sides:\[ \ln y = \ln x \]Now differentiate both sides with respect to \(x\):\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(\ln x) \]Using the derivative rule for \(\ln y\), \(\frac{1}{y} \frac{dy}{dx}\), and for \(\ln x\), \(\frac{1}{x}\), we have:\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \]

Solve for \( \frac{dy}{dx} \)

To find \( \frac{dy}{dx} \), multiply both sides by \( y \):\[ \frac{dy}{dx} = y \cdot \frac{1}{x} \]Since \( y = x \), we substitute:\[ \frac{dy}{dx} = x \cdot \frac{1}{x} = 1 \]

Key concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique used in calculus to differentiate functions that are products, quotients, or powers by taking the natural logarithm of the function first. By doing this, complex functions become much simpler and easier to differentiate. Here's how it works:

• Firstly, take the natural logarithm of both sides of the equation. This helps to transform the function into a form that is easier to work with.


• Differentiation of logarithms often simplifies multiplication or division into addition or subtraction, taking advantage of log properties.


• Once you have differentiated the log-transformed equation, use algebraic manipulation to solve for the derivative of the original function.

Using logarithmic differentiation can be particularly useful when dealing with functions that may not be straightforward to differentiate using standard rules. It's like breaking a complicated puzzle into smaller, manageable pieces.

Product Rule

The product rule is a fundamental differentiation rule used to find the derivative of a function that is the product of two or more functions. It states that:
If you have a function \( u(x) \cdot v(x) \), the derivative is given by:\[ (u \, v)' = u' \, v + u \, v' \]Where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively. Here’s how to apply it in steps:

• Differentiate the first function.


• Leave the second function as it is and multiply with the derivative of the first one.


• Do the reverse: leave the first function as it is and multiply with the derivative of the second one.


• Add the two results together, and you have the derivative of the product.

The product rule saves time and effort when working with products of multiple functions, ensuring you get the correct derivative each time.

Power Functions

Power functions are any functions of the form \( f(x) = x^n \) where \( n \) is a real number. Differentiating power functions is straightforward using the power rule in differentiation.
The power rule states:\[\frac{d}{dx}(x^n) = nx^{n-1}\]Here's how to apply the power rule:

• Take the exponent \( n \) and multiply it by the constant coefficient of the term.


• Reduce the exponent by 1.


• The result gives you the derivative of the power function.

In the original exercise, expressing functions in terms of powers simplifies subsequent steps, revealing how the power rule can directly simplify differentiation. When powers are combined through multiplication, adding the exponents as shown initially further streamlines calculation.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is approximately 2.71828. It is a widely used mathematical function in calculus for solving equations involving exponential growth and decay.
Key points about natural logarithms include:

• Logarithms transform products into sums and divisions into subtractions, which simplifies complex expressions during the differentiation process.


• The derivative of \( \ln(x) \) with respect to \( x \) is \( \frac{1}{x} \). This is particularly useful when applying logarithmic differentiation.


• Natural logarithms can also help solve equations where variables inside logarithm functions are too complex to isolate otherwise.

Understanding natural logarithms is essential in mastering calculus, as they often provide a clear pathway to solving otherwise difficult differentiation problems.