Question

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

Short answer

The derivative is \( h'(t) = \frac{2^t 3^t (\ln 2 - \ln 3) + 2 \cdot 2^t \ln 2 - 3 \cdot 3^t \ln 3}{(3^t + 2)^2} \).

Step-by-step solution

Identify the Structure

The given function is a quotient: \( h(t) = \frac{2^t + 3}{3^t + 2} \). This indicates that we need to use the quotient rule for differentiation, which states that if you have a function \( f(t) = \frac{u(t)}{v(t)} \), its derivative is given by \( f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} \). Here, \( u(t) = 2^t + 3 \) and \( v(t) = 3^t + 2 \).

Find Derivatives of the Numerator and Denominator

Differentiate \( u(t) = 2^t + 3 \). Since 3 is a constant, it differentiates to zero, and \( 2^t \) differentiates to \( 2^t \ln 2 \). Thus, \( u'(t) = 2^t \ln 2 \).Differentiate \( v(t) = 3^t + 2 \). Similarly, 2 is a constant which differentiates to zero, and \( 3^t \) differentiates to \( 3^t \ln 3 \). Thus, \( v'(t) = 3^t \ln 3 \).

Apply the Quotient Rule

Now that we have \( u(t) = 2^t + 3 \), \( u'(t) = 2^t \ln 2 \), \( v(t) = 3^t + 2 \), and \( v'(t) = 3^t \ln 3 \), we can apply the quotient rule:\[h'(t) = \frac{(2^t \ln 2)(3^t + 2) - (2^t + 3)(3^t \ln 3)}{(3^t + 2)^2}\]

Simplify the Expression

Distribute \( 2^t \ln 2 \) in the first term of the numerator and \( 3^t \ln 3 \) in the second term, then simplify:\[h'(t) = \frac{2^t \ln 2 \cdot 3^t + 2 \cdot 2^t \ln 2 - 2^t \cdot 3^t \ln 3 - 3 \cdot 3^t \ln 3}{(3^t + 2)^2}\]Combine similar terms:\[h'(t) = \frac{2^t 3^t \ln 2 - 2^t 3^t \ln 3 + 2 \cdot 2^t \ln 2 - 3 \cdot 3^t \ln 3}{(3^t + 2)^2}\]Factor out \( 2^t 3^t \) from the first two terms:\[h'(t) = \frac{2^t 3^t (\ln 2 - \ln 3) + 2 \cdot 2^t \ln 2 - 3 \cdot 3^t \ln 3}{(3^t + 2)^2}\]

Key concepts

Derivative

A derivative in calculus measures how a function changes as its input changes. It is a way to show the rate of change or the slope of the function at any given point. For a function of the form \( y = f(x) \), the derivative is represented as \( f'(x) \) or \( \frac{dy}{dx} \). This mathematical tool is essential because it provides insight into how different quantities are related and how they evolve, particularly in real-world applications such as physics and economics.
In our context, computing the derivative of \( h(t) = \frac{2^t + 3}{3^t + 2} \) helps us understand how the output \( h(t) \) changes as the input \( t \) changes. It tells us how sensitive the function is to small changes in \( t \). This understanding is crucial for analyzing trends and making predictions based on the behavior of the function.

Quotient Rule

The quotient rule is a method used to find the derivative of a ratio of two differentiable functions. It is expressed through the formula:

• If you have a function \( f(t) = \frac{u(t)}{v(t)} \), then the derivative \( f'(t) \) is given by:
  \[ f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} \]

This rule is especially handy when dealing with functions like \( h(t) = \frac{2^t + 3}{3^t + 2} \), where both the numerator \( u(t) \) and the denominator \( v(t) \) are functions of \( t \).
Knowing how to apply the quotient rule allows us to find the derivative of such functions efficiently. In essence, it involves calculating the derivatives of both the numerator and the denominator separately. Then, you plug those values into the quotient rule formula to find the desired derivative.

Exponential Functions

Exponential functions are a type of mathematical function where the variable is an exponent. They are written in the form \( a^t \), where \( a \) is a constant base and \( t \) is the exponent.

• In our problem, we have two exponential functions: \( 2^t \) and \( 3^t \).
• These functions grow (or decay) based on the base \( a \) – if \( a > 1 \), the function grows; if \( a < 1 \), it decays.

Understanding exponential functions is vital as they model real-world phenomena such as population growth, radioactive decay, and compound interest.
When differentiating exponential functions, the key is to remember that their derivative involves the original function scaled by the natural logarithm of the base. For example, the derivative of \( 2^t \) is \( 2^t \ln 2 \) and for \( 3^t \), it is \( 3^t \ln 3 \).

Differentiation

Differentiation is the process of finding a function's derivative. It's a fundamental operation in calculus and allows us to determine the rate of change of the function at any point.
This process involves applying rules of differentiation, like the power rule, product rule, or in our case, the quotient rule.

• We applied differentiation to find the derivatives of both \( u(t) = 2^t + 3 \) and \( v(t) = 3^t + 2 \).
• For \( u(t) \), differentiation yielded \( 2^t \ln 2 \). For \( v(t) \), it gave us \( 3^t \ln 3 \).

Differentiation transforms complex functions into simpler forms, making analyses and equations more manageable, especially when it involves determining the original function's behavior, trends, and slope.

Mathematical Notation

Mathematical notation is a language used to convey concepts of calculus and other mathematical fields succinctly.

• Notations like \( f'(t) \), \( u(t) \), and \( v(t) \) offer precise definitions of derivatives and functions.
• Key operators and symbols such as \( \ln \) (natural log) and \( \frac{dy}{dx} \) simplify complex ideas into comprehensible symbols.

Proper understanding of this notation is crucial for anyone working in mathematics or related fields, as it enables one to properly interpret and solve problems.
In our exercise, effective use of mathematical notation allowed us to apply differentiation correctly and formulate the solution for the derivative, showing each operation transparently and logically.