Problem 44 - Thomas Calculus

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. \(y=x^{\sin x}\)

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = x^{\sin x}(\cos x \cdot \ln x + \frac{\sin x}{x})\).

Step 1: Take the Natural Logarithm of Both Sides

Start by applying the natural logarithm to both sides of the equation. This gives us \(\ln y = \ln(x^{\sin x})\). By the properties of logarithms, we can bring the exponent down: \(\ln y = \sin x \cdot \ln x\).

Step 2: Differentiate Both Sides Implicitly

Differentiate both sides with respect to \(x\). For the left side, the derivative of \(\ln y\) is \(\frac{1}{y}\frac{dy}{dx}\). For the right side, use the product rule: the derivative of \(\sin x \cdot \ln x\) is \(\cos x \cdot \ln x + \sin x \cdot \frac{1}{x}\). So, the equation becomes \(\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \ln x + \frac{\sin x}{x}\).

Step 3: Solve for \(\frac{dy}{dx}\)

Multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = y(\cos x \cdot \ln x + \frac{\sin x}{x})\).

Step 4: Substitute Original Expression for \(y\)

We know from the given equation that \(y = x^{\sin x}\). Substitute this back in: \(\frac{dy}{dx} = x^{\sin x}(\cos x \cdot \ln x + \frac{\sin x}{x})\).

Key Concepts

Implicit Differentiation

Implicit differentiation is a useful technique in calculus when you need to find the derivative of a function that is not explicitly solved for one variable in terms of another. In our exercise, instead of explicitly expressing the derivative of a function, implicit differentiation helps us work through the problem without needing to solve for the dependent variable directly.

When dealing with equations where both sides involve a variable, like \(\ln y = \ln(x^{\sin x})\), we must differentiate each side with respect to the independent variable, which is often \(x\). One key principle here is recognizing how to differentiate functions with respect to a variable implicitly.

• Start by taking the derivative of both sides.
• Use the chain rule where necessary, especially when differentiating expressions like \(\ln y\).
• Remember, \(\frac{d}{dx}\ln y = \frac{1}{y}\frac{dy}{dx}\).

This approach lets us keep the dependent variable, \(y\), intact until the final steps, where we solve for \(\frac{dy}{dx}\). This method is powerful, particularly for functions like our original \(y = x^{\sin x}\), which don't easily separate into parts we can differentiate directly.

Product Rule

The product rule is essential when differentiating expressions where two functions are multiplied together. This rule states that if \(u(x)\) and \(v(x)\) are two functions, their derivative is \(u'(x)v(x) + u(x)v'(x)\). In the context of our solution, we encounter the product rule during the differentiation of the right side of the equation:

• For \(\ln y = \sin x \cdot \ln x\), you identify the product of two functions \(\sin x\) and \(\ln x\).
• Apply the product rule: the derivative of \(\sin x \cdot \ln x\) is calculated by finding \(\cos x \cdot \ln x\) and adding it to \(\sin x \cdot \frac{1}{x}\).

This method ensures that each component of the product is considered separately. Breaking down the expression using the product rule sheds light on how changes in one function affect the overall product, which is an intuitive way to handle complex expressions. When you apply this rule effectively, it becomes much clearer how different parts of the equation contribute to the overall changes in the function.

Exponential Functions

Exponential functions often involve expressions where a variable appears in the exponent. These functions grow rapidly and can be challenging to differentiate directly. For our exercise, \(y = x^{\sin x}\), the variable \(x\) is both the base and part of the exponent, making it a complex but typical scenario for logarithmic differentiation.

By taking the natural logarithm of both sides as a first step, we simplify the differentiation process. The natural logarithm effectively "drops" the exponent, allowing us to treat the complex exponential function in manageable parts. This property of the logarithm is why \(\ln(x^{\sin x}) = \sin x \cdot \ln x\) is a crucial transformation.

• The natural log simplifies exponential functions, making them easier to differentiate.
• Always remember that after differentiation, you substitute back the original function for accuracy.

Recognizing when to simplify exponential expressions with logarithms is a key skill in calculus. It allows us to handle equations involving exponential growth intuitively and accurately.