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Chapter 4: Problem 33

In Exercises 1 through \(34,\) find the derivative. $$ e^{x^{2}} \ln x $$

Short Answer

Expert verified
The derivative is \(\frac{e^{x^2}}{x} + 2xe^{x^2} \ln x\).

Step by step solution

01

Identify the Function

We need to find the derivative of the function \( f(x) = e^{x^2} \ln x \). This function is a product of two separate functions: \( u(x) = e^{x^2} \) and \( v(x) = \ln x \). To find the derivative, we'll employ the product rule.
02

Apply the Product Rule

The product rule states that if you have a function \( y = u(x) v(x) \), then its derivative \( y' \) is given by \( u'(x)v(x) + u(x)v'(x) \). We will need to find the derivatives \( u'(x) \) and \( v'(x) \).
03

Differentiate \( u(x) = e^{x^2} \)

To differentiate \( u(x) = e^{x^2} \), we use the chain rule. The derivative of \( e^u \) with respect to \( x \) is \( e^u \cdot u' \), where \( u = x^2 \). Thus, \( u'(x) = e^{x^2} \cdot 2x \).
04

Differentiate \( v(x) = \ln x \)

The derivative of \( \ln x \) with respect to \( x \) is \( \frac{1}{x} \). Therefore, \( v'(x) = \frac{1}{x} \).
05

Substitute into Product Rule

Using the values from previous steps, substitute into the product rule formula: \( f'(x) = e^{x^2} \cdot \frac{1}{x} + e^{x^2} \cdot 2x \cdot \ln x \).
06

Simplify the Expression

Simplify the expression obtained: \( f'(x) = \frac{e^{x^2}}{x} + 2xe^{x^2} \ln x \). This represents the derivative of the function \( e^{x^2} \ln x \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
The product rule is a fundamental tool in calculus for finding the derivative of the product of two functions. When you have two functions, say, \( u(x) \) and \( v(x) \), and you want to differentiate their product \( u(x)v(x) \), you can't simply differentiate each one individually. You need to use the product rule formula, which states: \[ (uv)' = u'v + uv' \] Here's how you apply it step-by-step:
  • Find the derivative of \( u(x) \), denote it as \( u'(x) \).
  • Find the derivative of \( v(x) \), denote it as \( v'(x) \).
  • Multiply \( u'(x) \) by \( v(x) \).
  • Multiply \( u(x) \) by \( v'(x) \).
  • Add the two products together to get the derivative of \( u(x)v(x) \).
Using the product rule ensures that no terms are overlooked in the derivation process, providing a comprehensive view of how both parts of the function contribute to the overall change.
Chain Rule
The chain rule is used in calculus when dealing with composite functions - that is, functions within other functions. If you have a function \( h(x) = f(g(x)) \), the chain rule helps you differentiate by taking the derivative of the outer function evaluated at the inner function and multiplying it by the derivative of the inner function. The formula for the chain rule is: \[ (f(g(x)))' = f'(g(x)) \, g'(x) \] Let's see how this works:
  • Identify the inner function \( g(x) \).
  • Identify the outer function \( f(u) \), where \( u = g(x) \).
  • Differentiate the outer function concerning \( u \), denoting it as \( f'(u) \).
  • Differentiate the inner function, denoting it as \( g'(x) \).
  • Substitute \( g(x) \) into the derivative of the outer function, then multiply by the derivative of the inner function.
The chain rule keeps everything linked, just like a chain, ensuring smooth computation of derivatives across nested functions.
Derivative of Logs
When you come across logarithmic functions in calculus, the derivative of the natural log function \( \ln x \) is particularly useful to remember. The derivative of \( \ln x \) with respect to \( x \) is quite simple: \[ \frac{d}{dx} \ln x = \frac{1}{x} \] Here's what this implies:
  • The logarithmic function grows slower as \( x \) increases, hence the \( \frac{1}{x} \) term, which gets smaller with increasing \( x \).
  • This derivative is crucial when solving equations involving logarithms, particularly when applying it as part of the product or chain rule.
Additionally, for other bases \( \ln_a x \), the derivative incorporates a scaling factor: \[ \frac{d}{dx} \ln_a x = \frac{1}{x \ln a} \] So remember, with \( \ln x \), the derivative simplifies calculations significantly when combined with other differentiation rules.
Exponential Functions
Exponential functions, denoted generally as \( e^{x} \), have a unique property: they are their own derivatives. This makes them relatively simple to differentiate:
  • For \( e^{u} \), where \( u = u(x) \), the derivative is \( e^{u} \cdot u' \).
  • The term \( u' \) arises from applying the chain rule within the exponential function.
This property is extremely useful because it simplifies the process of differentiation for functions that involve exponentials. Imagine dealing with \( e^{x^2} \). First identifying \( u = x^2 \), you apply the chain rule:
  • Differentiate \( u(x) \) to find \( u'(x) = 2x \).
  • The derivative, using the chain rule, is then \( e^{x^2} \cdot 2x \).
This pattern of differentiating exponentials frequently appears in calculus, highlighting their pervasive relevance and the elegance of their mathematics.

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