Problem 5 - Calculus Single Variable

Question

\( y\) is a function of \(x .\) Calculate the derivative of the given expression with respect to \(x\). (Your answer should contain the term \(d y / d x .)\) \(x e^{y}\)

Step-by-Step Solution

Answer

The derivative of \( x e^y \) with respect to \(x\) is \( e^y + x e^y \frac{dy}{dx} \).

Step 1: Identify Given Expression

The function provided is an equation involving two variables, given by the expression \( x e^{y} \), where \(y\) is a function of \(x\). We are asked to find its derivative with respect to \(x\).

Step 2: Apply Product Rule

To find the derivative of the expression \( x e^{y} \), apply the product rule, which states that \((uv)' = u'v + uv'\), where \(u = x\) and \(v = e^y\). First, compute the derivative of the terms separately.

Step 3: Derivative of the First Term

Calculate the derivative of \(u = x\) with respect to \(x\). The derivative of \(x\) is 1, so \(u' = 1\).

Step 4: Derivative of the Second Term

Now apply the chain rule to differentiate \(v = e^y\) with respect to \(x\). The derivative of \(e^y\) is \(e^y\cdot\frac{dy}{dx}\) because \(y\) is a function of \(x\). Thus, \(v' = e^y\cdot\frac{dy}{dx}\).

Step 5: Combine Results Using Product Rule

Substitute the derivatives into the product rule formula: \[\frac{d}{dx}(x e^y) = u'v + uv'\]Substitute \(u' = 1\), \(v = e^y\), \(u = x\), and \(v' = e^y\frac{dy}{dx}\):\[1 \cdot e^y + x \cdot e^y \frac{dy}{dx}\]Simplify to obtain the result.

Step 6: Compose Final Answer

The expression simplifies to:\[e^y + x e^y \frac{dy}{dx}\]This is the derivative of the given expression with respect to \(x\), containing the term \(\frac{dy}{dx}\).

Key Concepts

The Product Rule Explained

When dealing with derivatives, the product rule is an essential tool for finding the derivative of the product of two functions. Let's consider an expression that involves a product like in our exercise, namely, \( u = x \) and \( v = e^y \), where \( y \) is a function of \( x \). The product rule helps us differentiate such expressions.

The product rule is expressed mathematically as:

• \( (uv)' = u'v + uv' \)

It means, to find the derivative of \( uv \), we need to:

• Differentiate \( u \) with respect to \( x \) to get \( u' \)
• Differentiate \( v \) with respect to \( x \) to get \( v' \)
• Then apply the rule: \( u'v + uv' \)

In our expression \( xe^y \), we took the derivative of \( x \) as 1 and applied the chain rule to differentiate \( e^y \).

By using the product rule, we ensure that the change between the two parts is accounted for accurately, allowing us to compute the derivative of more complex interactions between the functions.

The Chain Rule Simplified

The chain rule is crucial when differentiating composite functions. A composite function consists of one function inside another, like \( e^y \) in our exercise, where \( y \) is a function of \( x \). The chain rule is built to handle such cases effectively by taking derivatives at each stage of the composition.

The chain rule can be stated as:

• If \( z = f(g(x)) \), then the derivative is \( \frac{dz}{dx} = f'(g(x)) \cdot g'(x) \)

Similarly, in our case, \( e^y \) can be seen as \( f(y) \) where \( y \) further depends on \( x \). Hence, the derivative becomes:

• \( \frac{d}{dx}(e^y) = e^y \cdot \frac{dy}{dx} \)

By applying the chain rule in our problem,

• we help consider how both the outer function \( e^y \) and its derivative \( \frac{dy}{dx} \) affect the overall differentiation

Remember, using the chain rule is essential whenever you differentiate functions nested within other functions.

Understanding Implicit Differentiation

Implicit differentiation is a technique used when an equation implicitly defines a function, meaning the function isn’t isolated on one side. In our exercise, \( y \) is not given explicitly as a function of \( x \) but implicitly through the expression \( x e^y \).

Sometimes, it’s tough to solve for \( y \) in terms of \( x \) outright. Implicit differentiation allows us to handle this by:

• Differentiating each term of the equation directly with respect to \( x \)
• Treating \( y \) as a function of \( x \) (that's why we use \( \frac{dy}{dx} \))
• Applying the derivative rules (like product rule and chain rule) as needed

For instance, in the expression \( x e^y \):

• We differentiate \( e^y \) by using the chain rule and add \( \frac{dy}{dx} \)
• The product rule helps differentiate the entire expression \( x e^y \)

In conclusion, implicit differentiation is useful when direct functions aren't straightforwardly expressed and gives us a powerful method to find derivatives of such expressions easily.