\[\begin{align*} \boxed{\frac{d}{dx} \left( 2x \ln(x) \right) = 2 \ln(x) + 2} \end{align*}\]
Introduction
To find the derivative of \( 2x \ln(x) \), we will use the product rule. Learning to work with logarithmic expressions is important in calculus as they show up later when tackling integration.
Step-by-Step Solution
The product rule, which states that if we have two functions \( u(x) \) and \( v(x) \), then:
\[\begin{align*} \frac{d}{dx} \left( u(x) \cdot v(x) \right) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \end{align*}\]
\noindent For our function \( 2x \ln(x) \):
– Let \( u(x) = 2x \)
– Let \( v(x) = \ln(x) \)
Now, we’ll differentiate each part separately.
Step 1: Differentiate \( u(x) = 2x \)
The derivative of \( u(x) = 2x \) is simply:
\[\begin{align*} u'(x) = 2 \end{align*}\]
Step 2: Differentiate \( v(x) = \ln(x) \)
The derivative of \( v(x) = \ln(x) \) is:
\[\begin{align*} v'(x) = \frac{1}{x} \end{align*}\]
Step 3: Apply the Product Rule
Now, substitute \( u(x) \), \( v(x) \), \( u'(x) \), and \( v'(x) \) into the product rule formula:
\[\begin{align*} \frac{d}{dx} \left( 2x \ln(x) \right) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \end{align*}\]
Substitute in the values:
\[\begin{align*} = 2 \cdot \ln(x) + 2x \cdot \frac{1}{x} \end{align*}\]
Simplify the expression:
\[\begin{align*} = 2 \ln(x) + 2 \end{align*}\]
Thus, the derivative of \( 2x \ln(x) \) is:
\[\begin{align*} \frac{d}{dx} \left( 2x \ln(x) \right) = 2 \ln(x) + 2 \end{align*}\]
Conclusion
To summarize, the derivative of \( 2x \ln(x) \) is \( 2 \ln(x) + 2 \). Using the product rule allowed us to handle the derivative efficiently. Mastering derivatives of expressions like this one helps make working with logarithmic and exponential growth not so scary.
The final answer is:
\[\begin{align*} \frac{d}{dx} \left( 2x \ln(x) \right) = 2 \ln(x) + 2 \end{align*}\]
