Derivative Calculation Using Chain Rule

This problem asks us to find the derivative \(\frac{dy}{dx}\) for a composite function. We are given two functions: \(y\) in terms of \(u\), and \(u\) in terms of \(x\). To solve this, we will use the chain rule, which is a fundamental concept in differential calculus.

Understanding the Chain Rule Formula

The chain rule is essential for differentiating functions that are composed of other functions. If \(y\) is a function of an intermediate variable \(u\), and \(u\) itself is a function of \(x\), then the derivative of \(y\) with respect to \(x\) is given by the product of their individual derivatives:

\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]

Let's apply this rule to the specific functions provided in the question.

Step-by-Step Derivative Process

We are given the following relationships:

• The outer function: \(y = u^2 + \log u\)
• The inner function: \(u = e^x\)

First, determine \(\frac{dy}{du}\):

We need to differentiate the function \(y = u^2 + \log u\) with respect to \(u\). We will differentiate each term separately:

• The derivative of \(u^2\) with respect to \(u\) is \(2u\).
• The derivative of \(\log u\) with respect to \(u\) is \(\frac{1}{u}\).

Combining these, we get:

\[ \frac{dy}{du} = \frac{d}{du}(u^2) + \frac{d}{du}(\log u) = 2u + \frac{1}{u} \]

Next, determine \(\frac{du}{dx}\):

Now, we differentiate the function \(u = e^x\) with respect to \(x\):

• The derivative of \(e^x\) with respect to \(x\) is simply \(e^x\).

So, we have:

\[ \frac{du}{dx} = \frac{d}{dx}(e^x) = e^x \]

Apply the Chain Rule to find the Final Derivative \(\frac{dy}{dx}\):

Now, we use the chain rule formula \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\) and substitute the derivatives we found:

\[ \frac{dy}{dx} = \left(2u + \frac{1}{u}\right) \times (e^x) \]

Substitute \(u = e^x\) into the expression for \(\frac{dy}{dx}\):

Since the final answer should be in terms of \(x\), we substitute \(u = e^x\) back into the expression for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \left(2(e^x) + \frac{1}{e^x}\right) \times e^x \]

Recall that \(\frac{1}{e^x}\) can be written as \(e^{-x}\). So the expression becomes:

\[ \frac{dy}{dx} = \left(2e^x + e^{-x}\right) \times e^x \]

Now, distribute \(e^x\) into the parenthesis:

\[ \frac{dy}{dx} = (2e^x \cdot e^x) + (e^{-x} \cdot e^x) \]

Using the exponent rule \(a^m \cdot a^n = a^{m+n}\):

\[ \frac{dy}{dx} = 2e^{x+x} + e^{-x+x} \]

\[ \frac{dy}{dx} = 2e^{2x} + e^0 \]

Since any non-zero number raised to the power of 0 is 1 (\(e^0 = 1\)):

\[ \frac{dy}{dx} = 2e^{2x} + 1 \]

This result can also be written as \(1 + 2e^{2x}\).

Matching with the Options

Let's compare our derived solution with the provided options:

Our calculated derivative, \(1 + 2e^{2x}\), perfectly matches Option 3.