Animated Solution for Mathematics - Differentiation: The derivative of ${\tan }^{-1}(\frac{\sin x-\cos x}{\sin x+\cos x})$, with respect to $x/2$, where $x\in (0,\pi /2)$ is :

Welcome, fellow traveler on the JEE journey! Today, we are tackling a problem that looks like a monster at first glance. We are asked to find the derivative of $f(x)={\tan }^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)$ with respect to $g(x)=x/2$.

If you try to jump straight into the derivative, you will find yourself drowning in a sea of quotient rules and chain rules. But hold on—take a deep breath. In mathematics, as in life, the most complex problems often have the most elegant solutions hidden just beneath the surface.

Unmasking the Trigonometric Identity

Let us look at the argument of our inverse tangent function: $\frac{\sin x-\cos x}{\sin x+\cos x}$. This structure is a classic signal. Whenever you see $\sin x$ and $\cos x$ added or subtracted, your first instinct should be to divide by $\cos x$.

Because $\frac{\sin x}{\cos x}=\tan x$ and $\frac{\cos x}{\cos x}=1$, this transforms our expression into:

Now, this looks familiar! It is almost the expansion of the compound angle formula $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$. To make it fit perfectly, we use the fact that $1=\tan (\pi /4)$.

We rewrite the expression as:

Suddenly, the 'monster' has vanished. We are left with $\tan (x-\pi /4)$.

The Domain Trap

Now, we substitute this back into our function: $f(x)={\tan }^{-1}(\tan (x-\pi /4))$. Many students will immediately cancel the ${\tan }^{-1}$ and $\tan$ to get $x-\pi /4$.

But wait! Stop! In the JEE, we never rush. We must check the domain. The problem gives us $x\in (0,\pi /2)$.

If we subtract $\pi /4$ from this interval, we get $(-\pi /4,\pi /4)$. Since this interval is strictly contained within the principal branch of the inverse tangent function $(-\pi /2,\pi /2)$, we are safe to simplify. Thus, $f(x)=x-\pi /4$.

The Final Calculus

Now, the problem has become trivial. We have $f(x)=x-\pi /4$ and $g(x)=x/2$. We need the derivative of $f$ with respect to $g$, which is $\frac{df}{dg}$.

Using the chain rule for parametric differentiation, we know that:

Calculating these derivatives is straightforward:

Dividing these gives us $1/(1/2)=2$. And there you have it! By looking past the complexity and finding the underlying structure, we have turned a terrifying calculus problem into a simple arithmetic result.

Keep this mindset—simplify first, differentiate later—and you will conquer any problem the JEE throws at you. The final answer is 2.