Problem 116 - Thomas Calculus in SI Units

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

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

Step 1: Apply Natural Logarithm

First, start by taking the natural logarithm of both sides of the equation. This will allow us to handle the right-hand side more easily.\[ \ln y = \ln (x^{\sin x}) \] Using the property of logarithms, we can rewrite this as:\[ \ln y = \sin x \cdot \ln x \]

Step 2: Differentiate Both Sides Implicitly

Now, differentiate both sides of the equation with respect to \(x\). For the left side, we have:\[ \frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx} \] For the right side, we apply the product rule. Differentiate \(\sin x\cdot \ln x\):\[ \frac{d}{dx}(\sin x\cdot \ln x) = (\cos x)\cdot \ln x + \sin x\cdot \frac{1}{x} \]

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

Set the expressions from Step 2 equal to each other:\[ \frac{1}{y} \cdot \frac{dy}{dx} = (\cos x)\cdot \ln x + \frac{\sin x}{x} \] Solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = y\left((\cos x)\cdot \ln x + \frac{\sin x}{x}\right) \]

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

Finally, substitute the original expression for \( y = x^{\sin x} \) into the formula obtained in Step 3:\[ \frac{dy}{dx} = x^{\sin x} \left( (\cos x)\cdot \ln x + \frac{\sin x}{x} \right) \]

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm with the base of \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. This concept is crucial in calculus whenever you need to simplify expressions, especially those involving exponentiation, and is widely used in logarithmic differentiation.In our problem, we applied the natural logarithm to both sides of the equation \( y = x^{\sin x} \). This transformation converts the complex power expression into a more manageable form: \( \ln y = \sin x \cdot \ln x \). Using the properties of logarithms, like \( \ln(a^b) = b \ln a \), helps deconstructing power terms into products, which are easier to differentiate.

Implicit Differentiation

Implicit differentiation is a technique in calculus used to find the derivative of a function that is not isolated on one side of the equation. It is particularly useful when dealing with equations that define \( y \) in terms of \( x \) in a less straightforward manner, where \( y \) cannot be neatly solved in terms of \( x \).For our equation \( \ln y = \sin x \cdot \ln x \), we implicitly differentiated both sides with respect to \( x \). For the left side, this gives \( \frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx} \). Here, \( y \) is treated as a function of \( x \), allowing us to find \( \frac{dy}{dx} \) even though \( y \) is not explicitly given as a function solely of \( x \).

Product Rule

The product rule is a fundamental rule in calculus used to differentiate products of two functions. If you have a product \( u(x) \cdot v(x) \), the derivative is:\[ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]In our differentiation process, the product rule helped us find the derivative of \( \sin x \cdot \ln x \). This was decomposed into two parts: the derivative of \( \sin x \) times \( \ln x \) plus \( \sin x \) times the derivative of \( \ln x \). Applying the product rule here gives:

• \( (\cos x) \cdot \ln x \) for differentiating \( \sin x \)
• \( \sin x \cdot \frac{1}{x} \) for differentiating \( \ln x \)

After applying the product rule, we get a combined and simplified expression for the derivative of the right-hand side of our equation.

Calculus

Calculus is a branch of mathematics focused on change and motion, exploring the rate of change (derivatives) and accumulation of quantities (integrals). Within this framework, we often deal with functions, their limits, and how they behave.Logarithmic differentiation, natural logarithms, implicit differentiation, and the product rule are all crucial techniques within calculus. These tools allow us to explore more complex relationships between variables, such as our power function \( y = x^{\sin x} \). Calculus provides the foundation to understand and solve problems involving rates of change, helping us uncover the behavior and properties of functions, whether simple or complex in nature.