Animated Solution for Mathematics - Differentiation: For all $x\in (0,1)$

A Journey into Calculus

Welcome, future engineers! Today, we are going to explore the elegant world of inequalities.

Often, when we see an expression like ${\log }_e(1+x)<x$ for $x\in (0,1)$, it might look simple, but it hides a profound truth about how functions behave. Let's embark on a journey to prove this rigorously.

The Intuition

Imagine you are standing on a graph. You have two paths: a straight line $y=x$ and a logarithmic curve $y={\log }_e(1+x)$.

Both paths start at the origin $(0,0)$. As you walk to the right, which path stays lower?

Visualizing this is the first step to mastery. We suspect the logarithmic curve stays below the line, but we must prove it for every single point in the interval $(0,1)$.

The Difference Function

In calculus, when we want to compare two functions, we create a new function that represents their difference.

Let's define:

If we can prove that $f(x)<0$ for all $x\in (0,1)$, we have effectively proven that ${\log }_e(1+x)<x$. It is a simple yet powerful transformation.

The Power of the Derivative

To understand how our difference function $f(x)$ behaves, we need to look at its rate of change. We differentiate $f(x)$ with respect to $x$:

Applying the chain rule, the derivative of ${\log }_e(1+x)$ is $\frac{1}{1+x}$, and the derivative of $x$ is $1$. Thus:

Let's simplify this expression:

The Conclusion

Now, look at the interval $x\in (0,1)$. In this range, $x$ is always positive.

Therefore, the numerator $-x$ is negative, and the denominator $1+x$ is positive. A negative divided by a positive is always negative, meaning $f^{\prime }(x)<0$ for all $x\in (0,1)$.

A negative derivative tells us that the function $f(x)$ is strictly decreasing. Since $f(0)={\log }_e(1)-0=0$, and the function is strictly decreasing, it must be that $f(x)<0$ for all $x>0$.

And there it is: ${\log }_e(1+x)-x<0$, which implies ${\log }_e(1+x)<x$. We have conquered the inequality!