Problem 120 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Take the Natural Logarithm of Both Sides

Start with the equation \( y = x^{\sin x} \). Apply the natural logarithm to both sides to facilitate differentiation, giving us \( \ln y = \ln(x^{\sin x}) \). Simplify the right side using the logarithmic identity \( \ln(a^b) = b \ln a \), resulting in \( \ln y = \sin x \cdot \ln x \).

Step 2: Differentiate Both Sides Implicitly

Differentiate both sides of the equation \( \ln y = \sin x \cdot \ln x \) with respect to \( x \). The derivative of the left side, using implicit differentiation, is \( \frac{1}{y} \cdot \frac{dy}{dx} \). For the right side, use the product rule: \( \frac{d}{dx}(\sin x \cdot \ln x) = \cos x \cdot \ln x + \frac{1}{x} \cdot \sin x \). This leaves us with the equation \( \frac{1}{y} \cdot \frac{dy}{dx} = \cos x \cdot \ln x + \frac{\sin x}{x} \).

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

Rearrange the expression to solve for \( \frac{dy}{dx} \) by multiplying both sides by \( y \). Thus, \( \frac{dy}{dx} = y \cdot (\cos x \cdot \ln x + \frac{\sin x}{x}) \).

Step 4: Replace \( y \) with \( x^{\sin x} \)

Since \( y = x^{\sin x} \), substitute this back into the equation to express \( \frac{dy}{dx} \) in terms of \( x \) only. This gives us \( \frac{dy}{dx} = x^{\sin x} \cdot (\cos x \cdot \ln x + \frac{\sin x}{x}) \).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations not easily expressed or solved for one variable in terms of the other. Instead of isolating the dependent variable, like in explicit differentiation, we differentiate with respect to the independent variable indirectly. In this process, we assume the dependent variable is a function of the independent variable, even if it's not stated directly. In the context of our exercise, when differentiating \( \ln y = \sin x \cdot \ln x \), we treat \( y \) as a function of \( x \), hence the derivative \( \frac{1}{y} \cdot \frac{dy}{dx} \). This allows us to seamlessly integrate derivative concepts without needing to explicitly solve for \( y \) first.

Product Rule

The product rule is an essential tool in calculus for finding the derivative of a product of two functions. It states that if you have two functions, \( u(x) \) and \( v(x) \), the derivative of their product \( u(x) \cdot v(x) \) is given by \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). This rule comes in handy when tackling more complicated expressions.

In our exercise, the product rule is applied to differentiate the expression \( \sin x \cdot \ln x \). We treat \( \sin x \) as \( u(x) \) and \( \ln x \) as \( v(x) \), making the derivative \( \cos x \cdot \ln x + \frac{1}{x} \cdot \sin x \). This illustrates how the product rule breaks down a complex differentiation task into manageable parts.

Natural Logarithm

The natural logarithm, denoted \( \ln \, \), is the logarithm to the base of the mathematical constant \( e \), approximately equal to 2.718. Natural logarithms simplify expressions with exponential qualities—particularly useful in logarithmic differentiation. In this context, the natural logarithm transforms power functions into products, simplifying differentiation.

For the exercise at hand, taking the logarithm of both sides of \( y = x^{\sin x} \) results in \( \ln y = \sin x \cdot \ln x \). This step leverages the property \( \ln(a^b) = b \ln a \), transforming the power into a multiplication, which is much easier to differentiate, as demonstrated in using implicit differentiation and the product rule.

Derivative Calculation

Derivative calculation involves finding how a function changes as its variables change. It is a cornerstone of calculus, used to determine the rate of change or slope of functions. In our exercise, the derivative of \( y = x^{\sin x} \) is sought using logarithmic differentiation.

Starting with \( \ln y = \sin x \cdot \ln x \), we differentiate both sides. Implicitly differentiating gives \( \frac{1}{y} \cdot \frac{dy}{dx} \) on the left. On the right, using the product rule, we find \( \cos x \cdot \ln x + \frac{\sin x}{x} \). Solving for \( \frac{dy}{dx} \), we rearrange, yielding \( \frac{dy}{dx} = y \cdot (\cos x \cdot \ln x + \frac{\sin x}{x}) \). Finally, substituting back, since \( y = x^{\sin x} \), results in \( \frac{dy}{dx} = x^{\sin x} \cdot (\cos x \cdot \ln x + \frac{\sin x}{x}) \), offering a clear view of the function's rate of change.