Problem 326 - Calculus Volume 1

Question

For the following exercises, compute \(d y / d x\) by differentiating \(\ln y\). $$y=x^{(e x)}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = x^{ex} (e \ln(x) + e) \).

Step 1: Take the natural logarithm

Start by applying the natural logarithm to both sides of the equation: \[ \ln(y) = \ln(x^{(ex)}) \]Using the property of logarithms \( \ln(a^b) = b\ln(a) \), rewrite the right-hand side as:\[ \ln(y) = ex \ln(x) \]

Step 2: Differentiate both sides with respect to x

Differentiate both sides of the equation with respect to \( x \):For the left side, use the chain rule:\[ \frac{d}{dx}(\ln(y)) = \frac{1}{y} \cdot \frac{dy}{dx} \]For the right side, use the product rule \(\frac{d}{dx}(uv) = u'v + uv'\) where \( u = ex \) and \( v = \ln(x) \):\[ \frac{d}{dx}(ex \ln(x)) = \frac{d}{dx}(ex) \cdot \ln(x) + ex \cdot \frac{d}{dx}(\ln(x)) \]which simplifies to:\[ e \ln(x) + \frac{ex}{x} \] therefore:\[ = e \ln(x) + e \]

Step 3: Solve for dy/dx

Setting the derivatives from Step 2 equal gives:\[ \frac{1}{y} \cdot \frac{dy}{dx} = e \ln(x) + e \]Isolate \( \frac{dy}{dx} \) by multiplying through by \( y \):\[ \frac{dy}{dx} = y(e \ln(x) + e) \]

Step 4: Substitute back the expression for y

Recall that \( y = x^{ex} \). Substitute this back into the expression for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = x^{ex} (e \ln(x) + e) \]Thus, the derivative \( \frac{dy}{dx} \) is:\[ x^{ex} (e \ln(x) + e) \]

Key Concepts

Chain Rule

In calculus, the chain rule is a fundamental tool used to differentiate composite functions. This rule is particularly useful when you have a function within another function. It provides a technique for finding the derivative of the outer function multiplied by the derivative of the inner function.

The formal expression of the chain rule is:\[\frac{d}{dx}[f(g(x))] = f'(g(x)) imes g'(x)\]In the original solution, the chain rule was applied when differentiating \( \ln(y) \). Since \( y \) itself is a function of \( x \), we need to consider its derivative, which results in \( \frac{1}{y} \times \frac{dy}{dx} \). This approach ensures all layers of the function are properly accounted for.

Product Rule

The product rule is another key differentiation rule in calculus. It comes into play when you need to differentiate the product of two functions. By using the product rule, you can systematically find the derivative of such expressions with ease.

The product rule is formulated as:\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]During the process of solving the original exercise, the product rule was used for differentiating \( ex \ln(x) \). If you identify \( u = ex \) and \( v = \ln(x) \), you apply the rule to get the derivative as a sum of two products. This was essential in getting the correct derivative for more complex functions.

Logarithmic Differentiation

Logarithmic differentiation offers a powerful method for differentiating functions that are too complex when tackled directly. This technique is especially useful for functions with variables in both the base and exponent.

Here's the basic idea:

• Start by taking the natural logarithm of both sides.
• Utilize properties of logarithms to simplify the expression.
• Differentiate both sides using rules like the chain rule and product rule.

In the given exercise, logarithmic differentiation helped in transforming \( y = x^{ex} \) to a more manageable form, \( \ln(y) = ex \ln(x) \), making differentiation more straightforward.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is approximately 2.71828. The natural logarithm simplifies the differentiation of certain functions due to its unique properties.

Some important points about \( \ln(x) \) are:

• The derivative \( \frac{d}{dx} \ln(x) \) is \( \frac{1}{x} \).
• It serves as an inverse function to the exponential function \( e^x \).
• It's frequently used in calculus for simplifying products and exponents.

In the exercise, \( \ln(y) \) helped express the power in \( x^{ex} \) as a product, which could then be differentiated conveniently using other differentiation rules.

Calculus

Calculus is the branch of mathematics that deals with the study of rates of change and accumulation. Two primary concepts govern calculus: differentiation and integration.

Differentiation focuses on computing derivatives of functions, which measure how a function changes as its input changes. In a broader sense, it helps in determining the slope of a curve, unlike algebra which deals with straight lines.

• It involves various rules like the chain rule and product rule, as described earlier.
• It provides tools to model dynamics systems and optimize complex functions.

The exercise revolves around differentiation, exploring techniques like the chain rule, product rule, and logarithmic differentiation, all cornerstones of calculus.