Problem 56 - Essential Calculus Early Transcendentals

Question

\(51-60=\) Use logarithmic differentiation or an alternative method to find the derivative of the function. $$ y=x^{\cos x} $$

Step-by-Step Solution

Answer

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

Step 1: Take the Natural Logarithm of Both Sides

To begin, take the natural logarithm (ln) of both sides of the equation. This will allow us to bring down the exponent:\[\ln y = \ln(x^{\cos x})\]Using the logarithm power rule, simplify as follows:\[\ln y = \cos(x) \cdot \ln(x)\]

Step 2: Differentiate Implicitly

Differentiate both sides of the equation with respect to \(x\):For the left side, we use implicit differentiation:\[\frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx}\]For the right side, we use the product rule:\[\frac{d}{dx}(\cos(x) \cdot \ln(x)) = \frac{d}{dx}(\cos x) \cdot \ln(x) + \cos(x) \cdot \frac{d}{dx}(\ln x)\]Thus, \[-\sin(x) \cdot \ln(x) + \cos(x) \cdot \frac{1}{x}\]

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

Combine the results from differentiating both sides and solve for \(\frac{dy}{dx}\):Substituting into the differentiated equation:\[\frac{1}{y} \cdot \frac{dy}{dx} = -\sin(x) \cdot \ln(x) + \frac{\cos(x)}{x}\]Multiply both sides by \(y\) to solve for \(\frac{dy}{dx}\):\[\frac{dy}{dx} = y \left( -\sin(x) \cdot \ln(x) + \frac{\cos(x)}{x} \right)\]

Step 4: Substitute Back for \(y\)

Replace \(y\) with the original function \(x^{\cos x}\):\[\frac{dy}{dx} = x^{\cos x} \left( -\sin(x) \cdot \ln(x) + \frac{\cos(x)}{x} \right)\]

Key Concepts

Implicit Differentiation

Implicit differentiation is a powerful method used when dealing with equations where the dependent variable cannot be easily isolated. Differentiating implicitly involves differentiating both sides of an equation with respect to a particular variable, typically \(x\),while treating other variables, like \(y\), as functions of \(x\).
In the given function \(y = x^{\cos x}\), we first take the natural logarithm of both sides to make the exponent manageable. Applying implicit differentiation to differentiate \(\ln y\) with respect to \(x\) leads us to:

• \(\frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx}\)

This process allows us to use known derivative rules to find \(\frac{dy}{dx}\), even when \(y\) is in an implicit form.

Product Rule

The product rule is essential when differentiating expressions where two functions are multiplied together. According to this rule, the derivative of a product \(u \cdot v\) is given by:

• \(\frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v'\)

In our case, while differentiating \(\cos(x) \cdot \ln(x)\), each component of the product requires its derivative:

• \(u = \cos(x)\), with \(u' = -\sin(x)\)
• \(v = \ln(x)\), with \(v' = \frac{1}{x}\)

Thus, applying the product rule involves computing:\(-\sin(x) \cdot \ln(x) + \cos(x) \cdot \frac{1}{x}\), capturing the combined effect of both components being differentiated.

Natural Logarithm

The natural logarithm, \(\ln\), is a fundamental mathematical function often used to simplify expressions involving exponents. Using \(\ln\) can transform complex powers into manageable multiplication forms.
In our example, applying \(\ln\) to \(y = x^{\cos x}\) allowed us to express:

• \(\ln(y) = \cos(x) \cdot \ln(x)\)

This transformation leverages the logarithmic power rule \(\ln(a^b) = b \cdot \ln(a)\), enabling easier differentiation. By doing this, the problem becomes much simpler to handle using implicit differentiation.

Exponent Manipulation

Manipulating exponents is a critical step when using logarithmic differentiation, particularly when dealing with non-linear expressions. The difficulty of differentiating a variable in the exponent is managed through logarithmic manipulation.
In \(y = x^{\cos x}\), directly handling the exponent \(\cos x\) complicates differentiation. However, by taking the natural logarithm of both sides, we transform it into a product: \(\cos(x) \cdot \ln(x)\).
This makes the task feasible, as dealing with a product in differentiation is considerably simpler compared to managing an exponent directly. With this solution, the differentiation becomes possible with known rules like the product rule, avoiding the complexity of directly tackling exponential structures.