Question

For the following exercises, compute $dy/dx$ by differentiating $\ln y$. $$y=x^e$$

Short answer

The derivative $\frac{dy}{dx}=e{x}^{e-1}$.

Step-by-step solution

Take the Natural Logarithm of Both Sides

Start by taking the natural logarithm (ln) of both sides of the equation. This will help us differentiate using the properties of logarithms.$$\ln y=\ln (x^e)$$

Simplify Using Logarithm Properties

Use the properties of logarithms to simplify the expression. The rule $\ln (a^b)=b\ln a$ can be applied here.$$\ln y=e\ln x$$

Differentiate Both Sides

Differentiate both sides of the equation with respect to $x$. We use implicit differentiation here.$$\frac{d}{dx}(\ln y)=\frac{d}{dx}(e\ln x)$$The derivative of $\ln y$ is $\frac{1}{y}\frac{dy}{dx}$, and the derivative of $e\ln x$ is $\frac{e}{x}$ using the constant multiple rule and the derivative of $\ln x$. Thus,$$\frac{1}{y}\frac{dy}{dx}=\frac{e}{x}$$

Solve for $\frac{dy}{dx}$

Multiply both sides of the equation by $y$ to isolate $\frac{dy}{dx}$.$$\frac{dy}{dx}=y\frac{e}{x}$$Since $y=x^e$, substitute $y$ back into the equation:$$\frac{dy}{dx}=x^e\frac{e}{x}$$

Simplify the Expression

Simplify the expression on the right side of the equation.$$\frac{dy}{dx}=e{x}^{e-1}$$

Key concepts

Natural Logarithm

The natural logarithm, denoted as $\ln$, is a logarithm with a base of the mathematical constant $e$ (approximately 2.71828). It is commonly used in calculus due to its unique properties and its connection to continuous growth rates. When you see $\ln y$, it represents the power to which $e$ must be raised to yield $y$. This concept is key in simplifying complex expressions and plays a crucial role during differentiation.
When we apply the natural logarithm to an equation, it can transform multiplication into addition and powers into multiplication, which are essential properties for simplifying expressions before differentiating. For example, taking the natural logarithm of $x^e$ lets us use the property $\ln (a^b)=b\ln a$, making differentiation more manageable.

Implicit Differentiation

Implicit differentiation is a technique used when a function is not explicitly solved for one variable. Instead of differentiating directly, we differentiate both sides of an equation with respect to a common variable.
In our exercise, the function $y=x^e$ is transformed by taking the natural logarithm into $\ln y=e\ln x$. By differentiating implicitly, we solve for $\frac{dy}{dx}$ even though $y$ might not be presented as a standalone function of $x$.

• Differentiate both sides with respect to $x$.
• Recognize that $\frac{d}{dx}(\ln y)=\frac{1}{y}\frac{dy}{dx}$, due to the chain rule.
• Continue to isolate $\frac{dy}{dx}$ by solving the resulting equation.

This technique is particularly useful in solving equations involving multiple variables that are intertwined.

Properties of Logarithms

The properties of logarithms are fundamental tools that simplify complex equations, particularly when differentiating. Here are the primary ones used:

• The Power Rule: $\ln (a^b)=b\ln a$ - This converts powers into a coefficient, simplifying differentiation.
• The Product Rule: $\ln (ab)=\ln a+\ln b$ - It simplifies multiplication within logarithms.
• The Quotient Rule: $\ln \left(\frac{a}{b}\right)=\ln a-\ln b$ - Useful for division.

In our exercise, we applied the power rule to transform $\ln (x^e)$ into $e\ln x$, streamlining the differentiation process. These logarithm properties are essential for both simplifying expressions and for applying derivative rules effectively.

Derivative Rules

Derivative rules form the backbone of calculus, allowing us to find rates of change. Key rules applied in this exercise include:

• The Derivative of the Natural Log: $\frac{d}{dx}(\ln x)=\frac{1}{x}$
• The Chain Rule: Essential when differentiating composite functions, where $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$.
• Constant Multiple Rule: $\frac{d}{dx}(cu)=c\frac{du}{dx}$ for any constant $c$.

In our solution, we used these rules to differentiate $\ln y$ and $e\ln x$, isolated $\frac{dy}{dx}$, and manipulated the original expression effectively through simplification. A solid understanding of these rules is necessary to tackle more complex differentiation problems.