Animated Solution for Mathematics - Differentiation: Find $\frac{dy}{dx}$ at $x=-1$, when $(\sin y{)}^{\sin \left(\frac{\pi x}{2}\right)}+\frac{\sqrt{3}}{2}{\sec }^{-1}(2x)+2^x\tan (\ln (x+2))=0$.

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of transcendental functions. We have an implicit equation:

Your goal is to find $\frac{dy}{dx}$ at $x=-1$. Take a deep breath; in the world of JEE Advanced, intimidation is the first trap. We will break this down into a sequence of elegant, manageable steps.

The Pre-Game Evaluation

Before we touch the derivative, we need to understand the state of our system at the specific point of interest, $x=-1$. Implicit differentiation often leaves us with terms involving $y$ that we must evaluate first.

Substituting $x=-1$ into our equation: 1. The first term becomes $(\sin y{)}^{\sin (-\frac{\pi }{2})}=(\sin y{)}^{-1}$. 2. The second term involves ${\sec }^{-1}(-2)=\frac{2\pi }{3}$. 3. The third term involves $\tan (\ln (-1+2))=\tan (0)=0$, causing the entire term to vanish.

We are left with the simplified relation:

Keep this value safe; it is our key to the final answer.

The Differentiation Battlefield

Now, we differentiate term by term. Let $T_1=(\sin y{)}^{\sin (\frac{\pi x}{2})}$. Using the logarithmic differentiation formula $\frac{d}{dx}(u^v)=u^v[\frac{v}{u}\frac{du}{dx}+\ln (u)\frac{dv}{dx}]$, we evaluate at $x=-1$.

Since the derivative of the exponent $v=\sin (\frac{\pi x}{2})$ involves $\cos (\frac{\pi x}{2})$, which is $0$ at $x=-1$, the expression simplifies significantly. The derivative of $T_1$ at $x=-1$ becomes:

Next, consider $T_2=\frac{\sqrt{3}}{2}{\sec }^{-1}(2x)$. Using the derivative rule $\frac{d}{dx}{\sec }^{-1}(u)=\frac{u^{\prime }}{|u|\sqrt{u^2-1}}$, we get:

At $x=-1$, this evaluates to $\frac{1}{2}$. Finally, for $T_3=2^x\tan (\ln (x+2))$, the product rule at $x=-1$ leaves us with $2^{-1}\cdot {\sec }^2(0)\cdot \frac{1}{0+1}=\frac{1}{2}$.

The Grand Assembly

We have arrived at the final assembly. The sum of our derivatives must be zero:

Rearranging for the derivative, we find:

Given $\sin y=-\frac{\sqrt{3}}{\pi }$, we use $\cos y=\pm \sqrt{1-{\sin }^{2}y}=\pm \sqrt{1-\frac{3}{{\pi }^2}}=\pm \frac{\sqrt{{\pi }^2-3}}{\pi }$. Substituting these into our expression, the final result is:

You have successfully navigated the complexity and arrived at the truth. This is the essence of mathematics—finding order in the apparent disorder.