Problem 45 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Suppose the function \(f\) satisfies the conditions: (i) \(f(x+y)=f(x) f(y)\) for all the \(x\) and \(y\) (ii) \(f(x)=1+x g(x)\) where \(\lim _{x \rightarrow 0} g(x)=1\). Show that the derivative \(f^{\prime}(x)\) exists and \(f^{\prime}(x)=f(x)\) for all \(x\).

Step-by-Step Solution

Answer

Using the given conditions, we derived that \(f^{\prime}(y) = f(y)\) for all x and y. We found the derivative \(f^{\prime}(x) = g(x) + xg^{\prime}(x)\) using the second condition. Since it was proven that \(f^{\prime}(y) = f(y)\) which includes \(f^{\prime}(x)\), we can finally conclude that the derivative \(f^{\prime}(x)\) exists and \(f^{\prime}(x) = f(x)\) for all x.

Step 1: Understand the functional equations given

We are given the following two conditions for the function f(x): (i) \(f(x+y) = f(x)f(y)\) for all x and y (ii) \(f(x) = 1 + xg(x)\) where \( \lim _{x \rightarrow 0} g(x)=1\)

Step 2: Utilize the properties of derivatives

We will start by finding the derivative of f(x). Note that the derivative of a sum is equal to the sum of the derivatives, so we can write \(f^{\prime}(x+y) = f^{\prime}(x) + f^{\prime}(y)\). To find the composite derivative, we will consider the first condition: \(f(x+y) = f(x)f(y)\) Now, differentiate both sides of the equation with respect to x: \(\frac{d}{dx}[f(x+y)] = \frac{d}{dx}[f(x)f(y)]\) Using chain rule for differentiation: \(f^{\prime}(x+y) = f^{\prime}(x)f(y) + f(x)f^{\prime}(y)\)

Step 3: Finding the derivative of f(x)

Now, consider the second condition: \(f(x) = 1 + xg(x)\) \(f^{\prime}(x) = \frac{d}{dx}[1+xg(x)] = g(x) + xg^{\prime}(x)\) where \(g^{\prime}(x)\) is the derivative of g(x).

Step 4: Substitution and solving

With the expressions for \(f^{\prime}(x+y)\) and \(f^{\prime}(x)\) at hand, consider the equation: \(f^{\prime}(x+y) = f^{\prime}(x)f(y) + f(x)f^{\prime}(y)\) Substitute \(x \rightarrow 0\) to the equation and use the properties of limits: \(f^{\prime}(y) = f^{\prime}(0)f(y) + f(0)f^{\prime}(y)\) Since \(f^{\prime}(x+y) = f^{\prime}(x) + f^{\prime}(y)\), we can assume that \(f^{\prime}(0)=0\). Thus, we can rewrite the previous equation as: \(f^{\prime}(y) = f(0)f^{\prime}(y)\) It's given that \( \lim _{x \rightarrow 0} g(x)=1\) which implies: \(f(0) = 1 + 0*g(0) = 1\) Thus, \(f^{\prime}(y) = f(y)\). Since we replaced x with y, we have \(f^{\prime}(x) = f(x)\) for all x.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It involves finding the rate at which a function changes at any given point. If you imagine a curve on a graph, differentiating the function represented by that curve helps you pinpoint the slope of the tangent line at any particular point. This is essential for understanding how variables change in relation to each other.If you have a function like \( f(x) = 1 + xg(x) \), knowing how to differentiate it allows us to compute \( f'(x) \) accurately. Here is how it works:

• When differentiating sums and products, remember the rules of differentiation, like the product rule and sum rule.
• Breaking down the function into manageable parts makes the differentiation easier.

Differentiation isn't just about applying rules mechanically. It requires understanding how these rules connect to the transformations of the original equation. For example, using \( f'(x) = g(x) + xg'(x) \) involves knowing what happens as the value of \( x \) becomes very small, linking to the concept of limits.

Limit of a Function

The concept of a limit is central to calculus and helps in understanding how functions behave as input values get very close to a certain point. Limits can describe the tendency of a function as a particular variable approaches a specific value. In this context, if you have a function \( f(x) = 1 + xg(x) \), with \( \lim _{x \rightarrow 0} g(x) = 1 \), the limit helps determine how \( g(x) \) affects \( f(x) \) when \( x \) is near 0.When you say a function approaches a limit, it means for every number close enough to this limit, the function eventually stays that close as \( x \) nears a particular value. Here's what to remember:

• Limits help in understanding the continuity and behavior of functions around specific points.
• They form the basis of defining the derivative, where the increment gets infinitesimally small.

By comprehending limits, you can understand how \( f(x) \) gets close to a certain value as \( x \) approaches zero which is essential for computing derivatives through a process known as differentiation.

Chain Rule

The Chain Rule is a valuable tool in differentiation when dealing with composite functions, or functions within other functions. It provides a systematic way of differentiating these more complex structures. If you have a scenario like the function three steps solution, where you differentiate \( f(x+y) = f(x)f(y) \), it becomes essential to use the Chain Rule.Let's break it down:

• The Chain Rule states that to differentiate a composite function \( h(x) = f(g(x)) \), you differentiate \( g(x) \) and then \( f(x) \), multiplying the results.
• In practice, it means taking one step at a time in breaking down the compound function into simpler parts.

For instance, differentiating \( f(x)f(y) \) involves the product rule in addition to the Chain Rule. By applying these rules step-by-step, you ensure that you accurately find the derivative of complex functions. Understanding how to utilize the Chain Rule effectively simplifies the process of differentiation, especially when functions become embedded within other functions.