Animated Solution for Mathematics - Differentiation: A function $f:R\to R$ satisfies the equation $f(x+y)=f(x)f(y)$ for all $x,y$ in $R$ and $f(x)\ne 0$ for any $x$ in $R$. Let the function be differentiable at $x=0$ and $f^{\prime }(0)=2$. Show that $f^{\prime }(x)=2f(x)$ for all $x$ in $R$. Hence, determine $f(x)$.

Welcome, fellow explorers of the mathematical universe. Today, we are not just solving a problem; we are uncovering the DNA of one of the most important functions in all of science: the exponential function.

You have been given a functional equation: $f(x+y)=f(x)f(y)$. At first glance, it looks like a simple algebraic rule, but it is actually a profound statement about how a system evolves. It tells us that the value of the function at a sum of two inputs is the product of the function at those individual inputs. This is the hallmark of exponential growth.

Finding the Anchor Point

Before we can use the heavy machinery of calculus, we need a starting point. We need to know where our function lives on the Cartesian plane. We are given $f(x+y)=f(x)f(y)$ and the condition that $f(x) eq 0$.

Let us test the simplest possible input: $x=0$ and $y=0$. Substituting these into our equation, we get $f(0+0)=f(0)f(0)$, which simplifies to $f(0)=[f(0){]}^2$.

Now, pause for a moment. We have a quadratic equation in terms of $f(0)$, which can be written as $f(0{)}^2-f(0)=0$, or $f(0)(f(0)-1)=0$. This gives us two possibilities: $f(0)=0$ or $f(0)=1$.

But wait! The problem explicitly states that $f(x) eq 0$ for any $x$. If $f(0)$ were $0$, then for any $x$, $f(x)=f(x+0)=f(x)f(0)=0$, which would make the function zero everywhere. Since that is forbidden, we must conclude that $f(0)=1$. We have found our y-intercept; this is our anchor.

The First Principle Microscope

Now, how do we find the derivative $f^{\prime }(x)$ when we don't have an explicit formula for $f(x)$? We go back to the absolute bedrock of calculus: the first principle of derivatives. The derivative is defined as the limit of the difference quotient as the interval $h$ shrinks to zero:

This formula is our microscope. It allows us to zoom in on the function at any point $x$ and see how it changes. But how do we handle $f(x+h)$? This is where the functional equation becomes our key. We know that $f(x+h)=f(x)f(h)$.

Let us substitute this into our limit:

Look at the numerator. We have a common factor of $f(x)$. Let us pull it out. Since the limit is only concerned with the variable $h$, $f(x)$ acts as a constant. We can move it outside the limit entirely:

The Birth of the Differential Equation

We are almost there. Look closely at that limit: ${\lim }_{h\to 0}\frac{f(h)-1}{h}$. Remember our anchor point? We know $f(0)=1$. So, we can replace the $1$ in the numerator with $f(0)$:

Does this look familiar? It is the definition of the derivative at $x=0$, denoted as $f^{\prime }(0)$. The problem tells us that $f^{\prime }(0)=2$. This is the slope of the tangent line at the y-intercept.

By substituting this value, our expression becomes:

Or, more elegantly, $f^{\prime }(x)=2f(x)$. This is a first-order linear differential equation. It tells us something profound: the rate of change of the function at any point is directly proportional to the value of the function at that point. This is the definition of exponential growth.

The Final Integration

To solve $f^{\prime }(x)=2f(x)$, we use the method of separation of variables. Let us write $f^{\prime }(x)$ as $\frac{df}{dx}$:

Now, we integrate both sides:

To find $C$, we use our anchor point $f(0)=1$. Substituting $x=0$ and $f(0)=1$:

So, we are left with $\ln (f(x))=2x$. To isolate $f(x)$, we exponentiate both sides:

And there it is. The function that transforms addition into multiplication is the exponential function. You have successfully derived the behavior of a system that grows in proportion to its current state. This is the math that describes everything from population growth to radioactive decay. You didn't just solve a problem; you derived a fundamental law of nature. Keep this intuition, and you will conquer any problem the JEE throws your way.