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Magazine Chapter 2: Problem 29
Compute the derivative of the given function. $$f(x)=\frac{3^{x^{2}}+x}{2^{x^{2}}}$$
Short Answer
Expert verified
Use the quotient rule and exponential differentiation to find the derivative.
Step by step solution
01
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 \( v(x) = 2^{x^2} \).
02
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, \( u'(x) = 3^{x^2} \cdot \ln(3) \cdot 2x + 1 \).
03
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, \( v'(x) = 2^{x^2} \cdot \ln(2) \cdot 2x \).
04
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 \( 2^{2x^2} \).
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
\[ f'(x) = \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{v(x)^2} \]
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.
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:
\[ \frac{d}{dx}[a^{g(x)}] = a^{g(x)} \ln(a) \cdot g'(x) \]
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, \(\frac{d}{dx}[2^{x^2}] = 2^{x^2} \ln(2) \cdot 2x\).
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:
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.
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 \(-x \cdot 2^{x^2} \cdot 2x \ln(2)\).
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.
