Problem 28 - A Textbook of Differential Calculus

Question

if \(2^{x}+2^{y}=2^{x+y}\), then \(\frac{d y}{d x}\) has the value equal to (a) \(-\frac{2^{y}}{2^{x}}\) (b) \(\frac{1}{1-2^{x}}\) (c) \(1-2^{y}\) (d) \(\frac{2^{x}\left(1-2^{y}\right)}{2^{y}\left(2^{x}-1\right)}\)

Step-by-Step Solution

Answer

The derivative \(\frac{dy}{dx}\) is equal to \(\frac{2^x(1-2^y)}{2^y(2^x-1)}\), which corresponds to option (d).

Step 1: Recognize the Given Expression

We start with the given equation: \(2^x + 2^y = 2^{x+y}\). Our goal is to find \(\frac{dy}{dx}\) from this equation.

Step 2: Simplify the Right Side

Recall the identity for adding exponents: \(2^a \cdot 2^b = 2^{a+b}\). Applying this, compare the given equality \(2^x + 2^y = 2^{x+y}\) with the knowledge that \(2^{x+y} = 2^x \cdot 2^y\).

Step 3: Rearrange the Equation

Since \(2^x + 2^y = 2^x \cdot 2^y\), rewrite it as \(2^x + 2^y - 2^x \cdot 2^y = 0\). This sets us up to find the derivative implicitly.

Step 4: Differentiate Implicitly

Differentiate the equation \(2^x + 2^y - 2^x \cdot 2^y = 0\) with respect to \(x\). Using implicit differentiation: 1. \(\frac{d}{dx}[2^x] = 2^x \ln(2)\),2. \(\frac{d}{dx}[2^y] = 2^y \ln(2) \frac{dy}{dx}\),3. \(\frac{d}{dx}[2^x \cdot 2^y] = (2^x \ln(2) 2^y) + (2^x 2^y \ln(2) \frac{dy}{dx})\). Substitute these into the derivative equation: \(2^x \ln(2) + 2^y \ln(2) \frac{dy}{dx} - (2^x \ln(2) 2^y + 2^x 2^y \ln(2) \frac{dy}{dx}) = 0\).

Step 5: Simplify the Derivative Equation

Rearrange the terms: - \(2^x \ln(2) + 2^y \ln(2) \frac{dy}{dx} - 2^{x+y} \ln(2) - 2^{x+y} \ln(2) \frac{dy}{dx} = 0\). Factorize to combine similar terms: - \((2^y \ln(2) \frac{dy}{dx} - 2^{x+y} \ln(2) \frac{dy}{dx}) + (2^x \ln(2) - 2^{x+y} \ln(2)) = 0\).Solve for \(\frac{dy}{dx}\), obtaining \(\ln(2) \frac{dy}{dx} (2^y - 2^{x+y}) = - 2^x \ln(2) (1 - 2^x)\).

Step 6: Solve for \(\frac{dy}{dx}\)

Isolate \(\frac{dy}{dx}\) by dividing both sides by the common factor \(\ln(2) (2^y - 2^{x+y})\), yielding:\[\frac{dy}{dx} = \frac{2^x (1 - 2^x)}{2^y - 2^{x+y}} = \frac{2^x \cdot (1 - 2^y)}{2^y \cdot (2^x - 1)}\].

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. In our context, the base is 2, and the exponents are variables like \(x\) and \(y\). These functions are vital in many fields, describing phenomena ranging from population growth to radioactive decay. When you see a function like \(2^x\), it represents an exponential growth where 2 is continually multiplied by itself according to the value of \(x\).
Understanding how exponential expressions add and multiply is crucial. For instance, the property \(2^{a+b} = 2^a \cdot 2^b\) means that if you multiply two exponential terms with the same base, you add their exponents. This property helps us simplify complex exponential equations and understand how they transform during calculations.

Implicit Differentiation Steps

Implicit differentiation is a technique used to find the derivative of an equation where \(y\) is not isolated on one side. This is often necessary when dealing with complex relations involving multiple variables. In our original question, \(2^x + 2^y = 2^{x+y}\), we need to differentiate both \(x\) and \(y\) without solving explicitly for \(y\).
The process involves several steps:

• Firstly, identify which parts of the equation need differentiation concerning \(x\).
• Apply the rules of differentiation to every term, treating \(y\) as a function of \(x\) (i.e., assume \(y = y(x)\)).
• When differentiating \(2^y\), remember to include \(\frac{dy}{dx}\), because \(y\) is dependent on \(x\).
• Combine and rearrange the resulting differentiated terms to isolate \(\frac{dy}{dx}\).

Implicit differentiation can be tricky initially, but with practice, it becomes a powerful tool for handling non-explicit relationships between variables.

Derivative Calculation

Calculating the derivative, especially when using implicit differentiation, requires step-by-step precision. Here, once we've applied differentiation to the equation \(2^x + 2^y - 2^{x+y} = 0\), we end up with several terms to simplify.

• The term \(2^x \ln(2)\) comes from differentiating \(2^x\) and serves as the standard derivative term for any exponential function \(a^x\), equivalent to \(a^x \ln(a)\).
• For the term \(2^y \ln(2) \frac{dy}{dx}\), it comes from differentiating \(2^y\), recognizing \(y\) as a function of \(x\).
• To solve for \(\frac{dy}{dx}\), rearrange the terms, factor common elements, and eventually isolate \(\frac{dy}{dx}\), resulting in the formula \(\frac{2^x (1 - 2^x)}{2^y - 2^{x+y}}\).

This process highlights the intricate balancing act required when applying implicit differentiation and showcases the connection between exponential growth and calculus.