Question

Compute the derivative of the given function. $$f(x)=2^{x^{3}+3 x}$$

Short answer

The derivative is \( f'(x) = 2^{x^3 + 3x} \ln(2) \times 3(x^2 + 1) \).

Step-by-step solution

Recognize the Function Type

The function given, \( f(x) = 2^{x^3 + 3x} \), is an exponential function where the base is a constant and the exponent is a function of \( x \). To find the derivative, we use the chain rule for differentiation and properties of exponential functions.

Start with the General Formula

For a function \( a^{u(x)} \), the derivative is \( \frac{d}{dx}\left(a^{u(x)}\right) = a^{u(x)} \ln(a) \frac{du}{dx} \). In our case, the base \( a \) is \( 2 \) and \( u(x) = x^3 + 3x \).

Differentiate the Exponent

Find the derivative of \( u(x) = x^3 + 3x \). This derivative \( \frac{du}{dx} = 3x^2 + 3 \), because the derivative of \( x^3 \) is \( 3x^2 \) and the derivative of \( 3x \) is \( 3 \).

Apply the Chain Rule

Substitute back into the formula: the derivative of \( f(x) = 2^{x^3 + 3x} \) is \( f'(x) = 2^{x^3 + 3x} \ln(2)(3x^2 + 3) \). This combines the exponential function's derivative rule and the chain rule.

Simplify the Expression

The expression \( f'(x) = 2^{x^3 + 3x} \ln(2) (3x^2 + 3) \) can be further simplified if required, recognizing that \( 3x^2 + 3 = 3(x^2 + 1) \). Thus, \( f'(x) = 2^{x^3 + 3x} \ln(2) \times 3(x^2 + 1) \).

Key concepts

Exponential Functions

Exponential functions are essential components in calculus and mathematical analysis. They are characterized by having a constant base raised to a variable exponent. In the function given, \( f(x) = 2^{x^3+3x} \), 2 is the constant base, and the expression \( x^3 + 3x \) is the variable exponent. Exponential functions have unique properties such as rapid growth or decay, which can be seen in real-world scenarios like population growth or radioactive decay.

• Rapid Growth: The value of an exponential function increases quickly as the variable in the exponent rises.
• Continuous Change: Exponential functions can model phenomena with continuous change.

Understanding their behavior provides insights into various dynamic systems. Typically, these functions exhibit exponential growth in mathematical terms of \( a^x \), where 'a' is a constant greater than one.

Chain Rule

The chain rule is a fundamental tool in calculus that deals with the differentiation of composite functions. Consider a function \( g(x) = f(h(x)) \) where you must differentiate a complex expression. The chain rule allows us to find the derivative of such functions by breaking them into simpler parts.
For the function \( f(x) = 2^{x^3 + 3x} \), the exponent \( x^3 + 3x \) acts as an inner function. The chain rule can be expressed as follows:

• Differentiate the outer function while keeping the inner function constant.
• Multiply by the derivative of the inner function.

In the context of our function, applying the chain rule involves differentiating \( 2^{x^3 + 3x} \) with respect to \( x \). First, treat \( 2^{u} \) as the outer function where \( u = x^3 + 3x \). Then differentiate \( u \) to find \( \frac{du}{dx} = 3x^2 + 3 \). The chain rule efficiently computes the derivative by systematically handling nested functions.

Derivative Calculation

Calculating the derivative is a core skill in calculus, and it's crucial for understanding how functions change at any given point. The derivative of a function measures its rate of change and is found using rules like the chain rule and basic differentiation principles.
For the function \( f(x) = 2^{x^3 + 3x} \), we first recognize it as an exponential function where the derivative follows the form \( \frac{d}{dx}(a^{u}) = a^{u} \ln(a) \cdot \frac{du}{dx} \). Here, the steps are:

• Identify \( a \) as 2 and \( u(x) \) as \( x^3 + 3x \).
• Find the derivative of \( u(x) \), which is \( 3x^2 + 3 \).
• Plug these into the formula yielding \( f'(x) = 2^{x^3 + 3x} \ln(2)(3x^2 + 3) \).
• Simplify further if necessary to \( f'(x) = 2^{x^3 + 3x} \ln(2) \times 3(x^2 + 1) \).

This process not only yields the derivative but also illustrates the systematic use of calculus rules to tackle complex functions.