Question

Use logarithmic differentiation to find \(\frac{d y}{d x}\). $$ y=x^{\sqrt{x}} $$

Short answer

\( \frac{dy}{dx} = \frac{x^{\sqrt{x}-0.5}(x\ln(x) + 2)}{2x} \).

Step-by-step solution

Take the natural logarithm of both sides

Starting with the equation \( y = x^{\sqrt{x}} \), take the natural logarithm of both sides to simplify the differentiation process. This gives us the equation \( \ln(y) = \ln(x^{\sqrt{x}}) \). Use the logarithmic identity \( \ln(a^b) = b\ln(a) \) to rewrite the right-hand side: \( \ln(y) = \sqrt{x} \cdot \ln(x) \).

Differentiate implicitly with respect to x

Differentiate both sides of the equation \( \ln(y) = \sqrt{x} \cdot \ln(x) \) with respect to \( x \). Use the chain rule on the left side to get \( \frac{1}{y} \frac{dy}{dx} \), and the product rule on the right side to give \( \frac{1}{2\sqrt{x}} \cdot \ln(x) + \sqrt{x} \cdot \frac{1}{x} \). Hence, the differentiated equation is \( \frac{1}{y} \frac{dy}{dx} = \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}} \cdot \frac{1}{x} \).

Solve for \( \frac{dy}{dx} \)

To solve for \( \frac{dy}{dx} \), multiply both sides by \( y \): \( \frac{dy}{dx} = y \left( \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{x\sqrt{x}} \right) \). Substitute back for \( y \) using \( y = x^{\sqrt{x}} \), giving us: \( \frac{dy}{dx} = x^{\sqrt{x}} \left( \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{x\sqrt{x}} \right) \).

Simplify the expression

Combine the terms inside the parentheses to achieve a single expression: \( \frac{ln(x)}{2\sqrt{x}} + \frac{1}{x\sqrt{x}} = \frac{x\ln(x) + 2}{2x\sqrt{x}} \). Therefore, the derivative is \( \frac{dy}{dx} = x^{\sqrt{x}} \cdot \frac{x\ln(x) + 2}{2x\sqrt{x}} \), which simplifies to \( \frac{dy}{dx} = \frac{x^{\sqrt{x}-0.5}(x\ln(x) + 2)}{2x} \).

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used when we deal with equations that are not easily solvable for one variable in terms of another. When an equation is expressed in terms of several functions and it's complex to isolate the variable we are differentiating with respect to, we apply implicit differentiation. Here, you're essentially differentiating each term with respect to the independent variable, mindful that all terms might involve the dependent variable either directly or indirectly.

For example, in the problem we're solving, we start with the equation given by \( y = x^{\sqrt{x}} \). When we take the natural logarithm (a step of logarithmic differentiation), our equation becomes \( \ln(y) = \sqrt{x} \cdot \ln(x) \). Implicit differentiation comes into play as we now need to differentiate each side of this equation with respect to \( x \). This means we look at \( \ln(y) \) as \( y \) implicitly dependent on \( x \), and differentiate accordingly, particularly using the chain rule for the left side and the product rule for the right.

• Using implicit differentiation involves handling derivatives of variables declared in terms of each other.
• For nested functions like \( \ln(y) \), remember to use the chain rule.

Chain Rule

The chain rule is an essential concept in calculus used to differentiate composite functions. A composite function is one where you have a function inside another function, such as \( f(g(x)) \). Instead of finding the derivative by expanding the function, the chain rule allows us to differentiate the outer function and multiply it by the derivative of the inner function.

In your exercise, while differentiating the equation \( \ln(y) = \sqrt{x} \cdot \ln(x) \), we handle the left side with the chain rule because \( y \) is a function of \( x \). The derivative of \( \ln(y) \) with respect to \( y \) is \( \frac{1}{y} \), but since \( y \) is dependent on \( x \), the chain rule directs us to multiply by \( \frac{dy}{dx} \), resulting in \( \frac{1}{y} \cdot \frac{dy}{dx} \). This illustrates perfectly how the chain rule operates.

• Recognize when a function is "inside" another function to decide when to use the chain rule.
• Differentiate the outer function first, then multiply by the derivative of the inner function.
• The chain rule is crucial for differentiating logarithmic, exponential, and trigonometric functions when nested.

Product Rule

The product rule is a tool for differentiating functions that are multiplied together. Given two functions \( u(x) \) and \( v(x) \), the product rule states that the derivative of their product is:\[(uv)' = u'v + uv'\]

In the logarithmic differentiation problem, within the expression \( \sqrt{x} \cdot \ln(x) \) on the right-hand side of the equation \( \ln(y) = \sqrt{x} \cdot \ln(x) \), we apply the product rule.Treat \( \sqrt{x} \) as \( u(x) \) and \( \ln(x) \) as \( v(x) \). Differentiating \( \sqrt{x} \) gives \( \frac{1}{2\sqrt{x}} \), and differentiating \( \ln(x) \) gives \( \frac{1}{x} \). Thus, using the product rule for differentiation, we obtain:\[(\sqrt{x} \cdot \ln(x))' = \frac{1}{2\sqrt{x}} \cdot \ln(x) + \sqrt{x} \cdot \frac{1}{x}\]

• The product rule is always used for two functions being multiplied.
• Differentiate each function independently, then use the formula.
• Ensure you use this technique whenever encountering products in differentiation tasks.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent, such as \( a^x \). Understanding how to differentiate exponential functions is essential, especially in equations like \( y = x^{\sqrt{x}} \), where exponentiation involves variables.

In this given problem, the expression \( x^{\sqrt{x}} \) illustrates an exponential function where both base and exponent involve the variable \( x \). When such functions are present, logarithmic differentiation simplifies the process. Taking the natural logarithm on both sides, we strategically transform our exponential function into a format allowing implicit differentiation. This technique is particularly useful when dealing with variable-powered variables or compound exponents.

• Exponentials can be tricky with variable bases and exponents, hence why logarithmic differentiation is used.
• Being comfortable converting exponentials via logarithms is crucial for this technique.
• Differentiation clarifies growth and decay patterns present in exponential functions.