Animated Solution for Mathematics - Differentiation: Let $f(x)=\frac{x}{\sqrt{a^2+x^2}}-\frac{d-x}{\sqrt{b^2+(d-x{)}^2}}$, $x\in R$ where $a,b$ and $d$ are non-zero real constants. Then:

Imagine you are standing before a complex, intimidating mountain of a problem. At first glance, the function

looks like a tangled mess of square roots and variables.

In the world of JEE Advanced, we do not fear complexity; we embrace it by breaking it down. The secret to solving this problem lies in the art of decomposition.

We do not need to solve the whole thing at once. Instead, let us split this function into two manageable parts:

By isolating these components, we can analyze their individual behaviors and then bring them back together for the grand finale.

Let us focus our energy on $g(x)=\frac{x}{\sqrt{a^2+x^2}}$. To find out if this part is increasing or decreasing, we need its derivative.

We use the quotient rule: the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

When we differentiate $g(x)$, we get:

After simplifying this expression—a moment where precision is key—we find that:

Since $a^2$ is always positive and the denominator is a positive power, $g^{\prime }(x)$ is strictly positive. This tells us that $g(x)$ is an increasing function. We have conquered the first peak!

Now, let us turn our attention to $h(x)=\frac{d-x}{\sqrt{b^2+(d-x{)}^2}}$. Notice the beautiful symmetry here? It is almost identical to $g(x)$, but with $b$ instead of $a$ and $(d-x)$ instead of $x$.

We can use the chain rule to differentiate this. Let $u=d-x$. Then $h(x)=\frac{u}{\sqrt{b^2+u^2}}$.

The derivative with respect to $u$ follows the same pattern we just derived:

However, we must not forget the chain rule! We must multiply by $\frac{du}{dx}$, which is the derivative of $(d-x)$ with respect to $x$. That derivative is $-1$.

So, we find:

This negative sign is the pivot point of the entire problem.

We are now ready to combine our findings. Our original function was $f(x)=g(x)-h(x)$. Therefore, the derivative is $f^{\prime }(x)=g^{\prime }(x)-h^{\prime }(x)$.

Substituting our results, we get:

Look at what happens! The two negative signs collide and transform into a positive sign. We are left with:

Both terms are strictly positive. The sum of two positive values is, without a doubt, positive. Thus, $f^{\prime }(x)>0$ for all real $x$.

The Beauty of Monotonicity

We have arrived at the summit. Because the first derivative $f^{\prime }(x)$ is strictly greater than zero for all real values of $x$, we can confidently conclude that $f(x)$ is a strictly increasing function.

This journey shows us that even the most daunting expressions are just collections of simpler, elegant patterns waiting to be revealed. Keep practicing, keep questioning, and remember: the math is always on your side if you take it one step at a time.