Question

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

Short answer

Use the quotient rule and exponential differentiation to find the derivative.

Step-by-step solution

Identify components for derivative

The function given is \( f(x) = \frac{3^{x^2} + x}{2^{x^2}} \), which is a quotient. To find the derivative, we'll use the quotient rule, which states that if \( f(x) = \frac{u(x)}{v(x)} \), then the derivative \( f'(x) \) is \( \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \). Here, \( u(x) = 3^{x^2} + x \) and \( v(x) = 2^{x^2} \).

Find derivatives of numerator and denominator components

First find the derivative \( u'(x) \). For \( 3^{x^2} \), use the formula \( \frac{d}{dx}[a^{g(x)}] = a^{g(x)} \cdot \ln(a) \cdot g'(x) \). So, let \( a = 3 \) and \( g(x) = x^2 \), then \( g'(x) = 2x \). Thus, \( \frac{d}{dx}[3^{x^2}] = 3^{x^2} \cdot \ln(3) \cdot 2x \). For \( x \), \( \frac{d}{dx}[x] = 1 \). So, \( u'(x) = 3^{x^2} \cdot \ln(3) \cdot 2x + 1 \).

Differentiate the denominator

Now, find the derivative \( v'(x) \) of \( v(x) = 2^{x^2} \) using the same exponential differentiation rule. So, \( \frac{d}{dx}[2^{x^2}] = 2^{x^2} \cdot \ln(2) \cdot 2x \). Thus, \( v'(x) = 2^{x^2} \cdot \ln(2) \cdot 2x \).

Apply the quotient rule

Now apply the quotient rule: \( f'(x) = \frac{(3^{x^2} \cdot \ln(3) \cdot 2x + 1) \cdot 2^{x^2} - (3^{x^2} + x) \cdot (2^{x^2} \cdot \ln(2) \cdot 2x)}{(2^{x^2})^2} \). Simplify the numerator by expanding the terms and combining like terms. The entire expression is over the common denominator \( 2^{2x^2} \).

Simplify the expression

Simplify the expression to get a clearer view of the derivative. The full derivative is:\[ f'(x) = \frac{3^{x^2} \cdot 2x \ln(3) \cdot 2^{x^2} + 2^{x^2} - (3^{x^2} + x) \cdot 2^{x^2} \cdot 2x \ln(2)}{4^{x^2}} \]Simplify the terms inside the fraction further if needed to provide a compact representation.

Key concepts

Quotient Rule

When dealing with derivatives of functions in the form \( \frac{u(x)}{v(x)} \), the **quotient rule** is the go-to tool. It allows us to find the derivative of these functions efficiently. The rule states:
\[ f'(x) = \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{v(x)^2} \]
In this exercise, we identify the numerator as \( u(x) = 3^{x^2} + x \) and the denominator as \( v(x) = 2^{x^2} \). When applying the quotient rule, remember:

• First, differentiate the numerator \( u(x) \) to find \( u'(x) \).
• Then, differentiate the denominator \( v(x) \) to find \( v'(x) \).
• Substitute these derivatives back into the quotient rule formula.

The result is a fraction with the difference of two products in the numerator, over the square of the denominator. This step ensures accuracy in finding the derivative of the original ratio.

Exponential Differentiation

When a function involves an expression like \(a^{g(x)}\), where \(a\) is a constant and \(g(x)\) is a function, we apply **exponential differentiation**. This involves using the chain rule along with the fact that the derivative of \(a^{g(x)}\) is:
\[ \frac{d}{dx}[a^{g(x)}] = a^{g(x)} \ln(a) \cdot g'(x) \]
In the problem, we encounter terms like \(3^{x^2}\) and \(2^{x^2}\). Here is how to approach these:

• For \(3^{x^2}\), consider \(a = 3\) and \(g(x) = x^2\). Calculate \(g'(x) = 2x\). Then use the formula to find \(\frac{d}{dx}[3^{x^2}] = 3^{x^2} \ln(3) \cdot 2x\).
• Similarly, for \(2^{x^2}\), set \(a = 2\) and \(g'(x) = 2x\). So, \(\frac{d}{dx}[2^{x^2}] = 2^{x^2} \ln(2) \cdot 2x\).

Exponential differentiation captures the growth characteristics of exponential functions and integrates into the larger calculation, such as the quotient rule.

Derivative Simplification

After applying the quotient rule, the expression of the derivative can appear complex and cluttered. **Derivative simplification** is about cleaning up this expression, making it easier to interpret and use in further mathematics.
First, focus on expanding and combining like terms in the numerator. You have:

• Combine terms like \(3^{x^2} \cdot 2x \ln(3) \cdot 2^{x^2}\) and \(-x \cdot 2^{x^2} \cdot 2x \ln(2)\).

Then, realize that the denominator \((2^{x^2})^2\) simplifies to \(4^{x^2}\).
Gather like terms and remove unnecessary variables to neaten the expression. The goal is to reach a form that is compact and conveys all required derivative information without extra complexity.