Question

Compute the derivative of the given function. $$m(w)=\frac{3^{w}+1}{2^{w}}$$

Short answer

The derivative of \( m(w) \) is \( \frac{3^{w} (\ln(3) - \ln(2)) - \ln(2)}{2^{w}} \).

Step-by-step solution

Identify the Quotient Rule

The function given is a quotient, in the form \( \frac{u(w)}{v(w)} \), where \( u(w) = 3^{w} + 1 \) and \( v(w) = 2^{w} \). To find its derivative, we will use the Quotient Rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).

Differentiate the Numerator

Find the derivative of \( u(w) = 3^{w} + 1 \). Using the rule for derivatives of exponential functions, we have \( u'(w) = 3^{w} \ln(3) \) since the derivative of a constant \(+1\) is zero.

Differentiate the Denominator

Find the derivative of \( v(w) = 2^{w} \). Similarly, using the rule for exponential functions, we get \( v'(w) = 2^{w} \ln(2) \).

Apply the Quotient Rule

Substitute \( u(w), u'(w), v(w), \) and \( v'(w) \) into the Quotient Rule formula: \[ m'(w) = \frac{3^{w} \ln(3) \cdot 2^{w} - (3^{w} + 1) \cdot 2^{w} \ln(2)}{(2^{w})^2} \]

Simplify the Expression

The expression can be simplified by factoring out common terms and simplifying the numerator and denominator: \[ m'(w) = \frac{3^{w} \cdot 2^{w} \ln(3) - 3^{w} \cdot 2^{w} \ln(2) - 2^{w} \ln(2)}{4^{w}} \]. Further simplification yields: \[ m'(w) = \frac{3^{w} 2^{w} (\ln(3) - \ln(2)) - 2^{w} \ln(2)}{4^{w}} \]. Extract \( 2^{w} \) from the numerator, giving: \[ m'(w) = \frac{2^{w} (3^{w}(\ln(3) - \ln(2)) - \ln(2))}{4^{w}} \]. Simplify further using \( 4^{w} = (2^{w})^2 \), leading to: \[ m'(w) = \frac{3^{w} (\ln(3) - \ln(2)) - \ln(2)}{2^{w}} \].

Finalize the Derivative

The derivative of the given function \( m(w)=\frac{3^{w}+1}{2^{w}} \) is: \[ m'(w) = \frac{3^{w} (\ln(3) - \ln(2)) - \ln(2)}{2^{w}} \].

Key concepts

Quotient Rule

When dealing with derivatives of functions represented as a quotient, the Quotient Rule becomes essential. It's used to differentiate functions of the form \( \frac{u}{v} \) where both \( u \) and \( v \) are functions of a variable, such as \( w \). The Quotient Rule formula is \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). This rule lets us find the derivative by first calculating the derivatives of the numerator and denominator individually. We then use these derivatives to compute the whole expression. Keep in mind that numerical steps must be handled with care for correct application. It’s important to simplify the derivative at the end of your calculation to achieve an easier-to-read solution.

Exponential Functions

Exponential functions are types of mathematical expressions where variables appear as exponents. An example is \( 3^w \) or \( 2^w \), where the base is a constant, and \( w \) is the exponent. When differentiating these functions, remember that they don't follow the power rule since the variable is in the exponent, not the base. The derivative of an exponential function \( a^w \) is \( a^w \ln(a) \), where \( \ln \) represents the natural logarithm. Thus, for \( 3^w \), its derivative is \( 3^w \ln(3) \). Similarly, \( 2^w \) gives a derivative as \( 2^w \ln(2) \). Mastery of logarithm properties enhances understanding and efficiency in finding these derivatives.

Simplification

Once the derivative is found using the Quotient Rule, simplification is key to expressing the result concisely. Simplification here involves combining like terms and reducing expressions for clarity. For instance, in our derivative example, the numerator and denominator were processed to factor out common terms like \( 2^w \). Additional steps, such as recognizing that \( 4^w \) can be written as \( (2^w)^2 \), aid in further reducing the complexity of the expression. Simplification is a crucial skill because it makes equations easier to handle and interpret, especially when checking for errors or alternative expressions.

Differentiation

Differentiation is the process of finding the derivative of a function. It tells us the rate at which a function is changing at any point. When working with functions involving exponential terms and quotients, differentiation provides insight into how variables interact to change the function's overall value. The key is to apply rules correctly, such as the Quotient Rule or exponential function differentiation. This helps us represent subtle changes in mathematical expression as numerical rates, essential for numerous applications like physics, economics, or engineering. Familiarity with these concepts increases proficiency in interpreting and solving complex real-world problems.