Chapter 3: Problem 39
Can the functions be differentiated using the rules developed so far? Differentiate if you can; otherwise, indicate why the rules discussed so far do not apply. $$y=\frac{2^{x}}{x}$$
Short Answer
Expert verified
Yes, it can be differentiated using the Quotient Rule.
Step by step solution
01
Identify the Function
Examine the function \( y = \frac{2^x}{x} \). This is a quotient of two functions: the numerator is an exponential function \( 2^x \) and the denominator is \( x \), a polynomial function.
02
Choose the Differentiation Rule
Since the function is a quotient of two differentiable functions, the Quotient Rule is appropriate for differentiation. Recall that the Quotient Rule states \( \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u = 2^x \) and \( v = x \).
03
Differentiate the Numerator
Differentiate the exponential function \( u = 2^x \) with respect to \( x \). Since the derivative of \( a^x \) is \( a^x \ln(a) \), we have \( u' = 2^x \ln(2) \).
04
Differentiate the Denominator
Differentiate the polynomial function \( v = x \) with respect to \( x \). The derivative of \( x \) is simply \( v' = 1 \).
05
Apply the Quotient Rule
Substitute \( u, u', v, \) and \( v' \) into the Quotient Rule formula: \[ \left( \frac{2^x}{x} \right)' = \frac{x \cdot 2^x \ln(2) - 2^x \cdot 1}{x^2} = \frac{x \cdot 2^x \ln(2) - 2^x}{x^2}. \]
06
Simplify the Result
Factor out \( 2^x \) from the numerator: \[ = \frac{2^x (x \ln(2) - 1)}{x^2}. \] This is the differentiated form of the original function. Both the numerator and denominator were differentiable using standard rules.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The quotient rule is a handy tool in calculus when dealing with the division of two differentiable functions. Imagine you have a quotient, a division, of two functions such as \( \frac{u}{v} \). To differentiate this expression, we must use the quotient rule. The quotient rule formula is:
- \[ \left( \frac{u}{v} \right)' = \frac{v \cdot u' - u \cdot v'}{v^2} \]
- First, differentiate the numerator \(u\) to get \(u'\).
- Next, differentiate the denominator \(v\) to get \(v'\).
- Then, apply the products as shown in the formula.
Exponential Functions
Exponential functions are a class of functions that include a constant base raised to a variable exponent, such as \( 2^x \), where 2 is the base and \(x\) is the exponent.These types of functions grow rapidly and have some distinct features:
- The base of an exponential function is a positive constant.
- The exponent is the variable; it changes.
- The function is smooth and continuous.
- \( a^x \ln(a) \)
Polynomial Functions
Polynomial functions are among the most common types of functions encountered in calculus, defined as sums of constants multiplied by variables raised to non-negative integer powers. A simple polynomial function is \( x^n \), where \( n \) is a non-negative integer.The polynomial hierarchy by degree includes:
- Linear functions such as \( x \), which are first-degree polynomials.
- Quadratic functions, like \( x^2 \), which are second-degree polynomials.
- Higher-degree polynomials, such as \( x^3 \) and beyond.
- \( \text{If } f(x) = x^n, \text{ then } f'(x) = n x^{n-1} \)