Problem 82 - Calculus and its applications

Question

Differentiate. $$ g(x)=e^{2 x} \ln x $$

Step-by-Step Solution

Answer

The derivative is \( g'(x) = 2e^{2x} \ln x + \frac{e^{2x}}{x} \).

Step 1: Identify the Type of Function

The function given is a product of two functions: \( e^{2x} \) and \( \ln x \). This requires the use of the product rule for differentiation.

Step 2: Recall the Product Rule

The product rule states that if you have two functions \( u(x) \) and \( v(x) \), their derivative is given by \( u'(x)v(x) + u(x)v'(x) \). Let's identify \( u(x) = e^{2x} \) and \( v(x) = \ln x \) for our function.

Step 3: Differentiate \( u(x) = e^{2x} \)

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

Step 4: Differentiate \( v(x) = \ln x \)

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

Step 5: Apply the Product Rule

Substitute the derivatives into the product rule formula: \( g'(x) = u'(x)v(x) + u(x)v'(x) \). Substitute \( u'(x) = 2e^{2x} \), \( u(x) = e^{2x} \), \( v'(x) = \frac{1}{x} \), and \( v(x) = \ln x \).

Step 6: Calculate the Expression

Using the product rule, \[ g'(x) = 2e^{2x} \ln x + e^{2x} \cdot \frac{1}{x} \]. Simplify this expression: \[ g'(x) = 2e^{2x} \ln x + \frac{e^{2x}}{x} \].

Key Concepts

Product Rule

The Product Rule is a fundamental technique in calculus for differentiating products of two functions. When you have two functions, say \( u(x) \) and \( v(x) \), and you need to find the derivative of their product, the product rule is applied. The rule states that the derivative of \( u(x) \cdot v(x) \) is given by \( u'(x)v(x) + u(x)v'(x) \).

• \( u'(x) \) is the derivative of \( u(x) \).
• \( v'(x) \) is the derivative of \( v(x) \).

In our example, \( g(x) = e^{2x} \ln x \), we treat \( e^{2x} \) as \( u(x) \) and \( \ln x \) as \( v(x) \). First, find the derivatives \( u'(x) = 2e^{2x} \) and \( v'(x) = \frac{1}{x} \). Then, apply the product rule formula. This results in:\[g'(x) = 2e^{2x} \ln x + \frac{e^{2x}}{x}\]The product rule ensures that each component of this multiplication is accounted for when carrying out differentiation.

Chain Rule

The Chain Rule is used when differentiating a composite function, a situation where one function is nested within another. If you have a function \( h(x) = f(g(x)) \), the derivative is found by multiplying the derivative of the outer function by the derivative of the inner function. Mathematically, it's expressed as:\[h'(x) = f'(g(x)) \cdot g'(x)\]In our scenario, when differentiating \( e^{2x} \), the chain rule is necessary. Here, \( f(x) = e^x \) and \( g(x) = 2x \). Differentiating \( f(x) \) gives \( f'(x) = e^x \) and differentiating \( g(x) \) gives \( g'(x) = 2 \). Applying the chain rule, we multiply these derivatives:\[u'(x) = f'(g(x)) \cdot g'(x) = e^{2x} \cdot 2 = 2e^{2x}\]This method allows us to find the derivative of functions that might otherwise seem complex at first glance.

Exponential Function

An exponential function in the context of calculus is usually of the form \( e^x \), where \( e \) is a constant approximately equal to 2.71828. It's a unique function where the rate of growth is proportional to the value of the function itself. One of the key properties of exponential functions is their simplicity in differentiation.

• The derivative of \( e^x \) is simply \( e^x \).
• When the exponent is more complex, such as \( e^{2x} \), the chain rule is employed for the derivative.

This makes \( e^x \) particularly straightforward to handle in differentiation compared to other functions that change form more dramatically. In our exercise, with the function \( e^{2x} \), you can see how the derivative combines with the chain rule to become \( 2e^{2x} \). Exponential functions are powerful in modeling growth processes and are frequent in numerous scientific disciplines.

Logarithmic Function

Logarithmic functions, especially the natural logarithm denoted as \( \ln x \), are pivotal in calculus due to their inversion relationship with exponential functions. The natural logarithm is the inverse of the exponential function \( e^x \).

• The derivative of \( \ln x \) is \( \frac{1}{x} \).
• This characteristic makes logarithms advantageous for simplifying complex expressions, especially when differentiating products or quotients.

In the example function \( g(x) = e^{2x} \ln x \), the straightforward derivative of \( \ln x \) helps simplify the application of the product rule. Logarithmic functions unpack complexities of growth and decay, helping simplify derivative calculations by converting multiplicative relationships into additive ones, aiding in easier manipulation and solution derivations.