Problem 51 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated derivative. \(f^{\prime}(1)\) if \(f(x)=x^{\sin x}\)

Step-by-Step Solution

Answer

\(f'(1) = \sin 1\).

Step 1: Express the Function Using Logarithm

The function is given as \( f(x) = x^{\sin x} \). To find the derivative, first express the function using a logarithm. Take the natural logarithm of both sides: \( \ln(f(x)) = \ln(x^{\sin x}) \). By applying the logarithm power rule, this becomes \( \ln(f(x)) = \sin x \cdot \ln x \).

Step 2: Differentiate Using Implicit Differentiation

Differentiate both sides with respect to \( x \). For the left-hand side, apply the chain rule: \( \frac{d}{dx}[\ln(f(x))] = \frac{1}{f(x)} \cdot f'(x) \). For the right-hand side, use the product rule: \( \frac{d}{dx}[\sin x \cdot \ln x] = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \).

Step 3: Solve for the Derivative \( f'(x) \)

Equating the derivatives from Step 2, we get: \( \frac{1}{f(x)} \cdot f'(x) = \cos x \cdot \ln x + \frac{\sin x}{x} \). Multiply both sides by \( f(x) \) to solve for \( f'(x) \): \( f'(x) = f(x) \cdot (\cos x \cdot \ln x + \frac{\sin x}{x}) \).

Step 4: Substitute \( f(x) = x^{\sin x} \) Back In

Since \( f(x) = x^{\sin x} \), substitute \( f(x) \) back into the expression for \( f'(x) \): \( f'(x) = x^{\sin x} \cdot (\cos x \cdot \ln x + \frac{\sin x}{x}) \).

Step 5: Evaluate the Derivative at \( x = 1 \)

Substitute \( x = 1 \) into the expression for \( f'(x) \): \( f'(1) = 1^{\sin 1} \cdot (\cos 1 \cdot \ln 1 + \frac{\sin 1}{1}) \). Simplify: \( 1^{\sin 1} = 1 \) and \( \ln 1 = 0 \), so \( f'(1) = 1 \cdot (0 + \sin 1) = \sin 1 \).

Key Concepts

Logarithms

Logarithms are an essential mathematical tool that helps in working with exponents, especially when dealing with complex functions. At its core, a logarithm is the inverse operation to exponentiation. For example, if you have a function expressed in the form of \( a^b \), taking the logarithm can simplify this by turning it into a multiplication problem: \( b \cdot \log(a) \). This can be incredibly helpful in calculus since multiplication is easier to differentiate than power functions. By transforming a function using logarithms, you can apply the rules of differentiation more straightforwardly.

• Natural Logarithm: Notated as \( \ln(x) \), the natural logarithm uses the base \( e \), a constant approximately equal to 2.718.
• Logarithm Power Rule: This rule states that \( \log(a^b) = b \cdot \log(a) \), which is highly useful when variables are in the exponent.

In the provided exercise, \(x^{\sin x}\) was rewritten using a natural logarithm as \(\ln(f(x)) = \sin x \cdot \ln x\), making it simpler to differentiate.

Implicit Differentiation

Implicit differentiation is a powerful technique in calculus used particularly for functions not solved explicitly for one variable in terms of the other. It allows us to compute derivatives without having to alter the form of the function, which can be crucial when dealing with more complex relationships.
When differentiating implicitly, you differentiate each term independently, much like how you differentiate in explicit terms, but apply the chain rule while considering each function part.
In our step-by-step solution:

• We applied the chain rule to differentiate \( \ln(f(x)) \), leading to the expression \( \frac{1}{f(x)} \cdot f'(x) \).
• This technique paired the logarithmic differentiation with the implicit method effectively, providing the derivative in a simplified form.

By applying these rules, you can analyze and derive complex functions that might not easily comply with regular derivative rules.

Product Rule

The product rule is essential 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:\[\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)\].This rule must be applied when both functions are dependent on the variable you are differentiating with respect to.
In the given problem, applying the product rule to \( \sin x \cdot \ln x \) involves:

• Finding the derivative of \( \sin x \), which is \( \cos x \).
• Finding the derivative of \( \ln x \), which is \( \frac{1}{x} \).
• Combining these results as: \( \cos x \cdot \ln x + \sin x \cdot \frac{1}{x} \).

By using the product rule correctly, the derivative inside the implicit differentiation was simplified, allowing for the comprehensive derivative to be computed efficiently.