Animated Solution for Mathematics - Differentiation: Let $f(x)=2x+{\tan }^{-1}x$ and $g(x)={\log }_e(\sqrt{1+x^2}+x),x\in [0,3]$. Then

Imagine you are standing at the starting line of a race. You have two runners, $f(x)$ and $g(x)$, both poised at the origin $(0,0)$.

Your goal is to determine which runner will be further ahead as they move along the track from $x=0$ to $x=3$. We are given the functions:

To understand their journey, we must look at their velocities—their derivatives.

Let us first analyze $f(x)$. Differentiating $f(x)$ is straightforward:

Since $x^2$ is always non-negative, the term $\frac{1}{1+x^2}$ is always between $0$ and $1$. Therefore, $f^{\prime }(x)$ is always between $2$ and $3$. This runner is moving quite fast!

Now, let us turn to $g(x)$. We apply the Chain Rule, where the derivative of $\ln (u)$ is $\frac{1}{u}\cdot u^{\prime }$. Here, $u=\sqrt{1+x^2}+x$, and the derivative of the inside part is $u^{\prime }=\frac{x}{\sqrt{1+x^2}}+1$.

When we multiply these, we get:

Look closely—the term $(\sqrt{1+x^2}+x)$ appears in both the numerator and the denominator. They cancel out, leaving us with the elegant result:

Now, we compare the velocities. For $x\in [0,3]$, $g^{\prime }(x)=\frac{1}{\sqrt{1+x^2}}$ is at most $1$ (when $x=0$) and at least $\frac{1}{\sqrt{10}}\approx 0.316$.

Meanwhile, $f^{\prime }(x)$ is always at least $2$. It is clear that $f^{\prime }(x)>g^{\prime }(x)$ for the entire interval.

Because both functions start at the same point, $f(0)=g(0)=0$, and $f(x)$ has a strictly higher rate of growth than $g(x)$, $f(x)$ must remain above $g(x)$ for all $x>0$.

Finally, we consider the maximum values. Since both functions are strictly increasing, their maximums occur at the rightmost boundary, $x=3$.

Because $f(x)$ has been growing faster and started at the same point, it is mathematically inevitable that:

This confirms our conclusion. By analyzing the derivatives, we have mapped the behavior of these functions across the entire domain. Always look for the rate of change, and the geometry of the problem will reveal itself to you.