Animated Solution for Mathematics - Differentiation: The equation of the normal to the curve $y=(1+x{)}^{2y}+{\cos }^2({\sin }^{-1}x)$ at $x=0$ is:

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are dissecting the anatomy of a curve.

We are given the equation $y=(1+x{)}^{2y}+{\cos }^2({\sin }^{-1}x)$ and tasked with finding the equation of the normal at $x=0$. It looks intimidating, but in the JEE Advanced arena, the most complex-looking problems often hide the most elegant solutions.

Before we touch a single derivative, let us look at the term ${\cos }^2({\sin }^{-1}x)$. Many students will immediately reach for the chain rule, but a seasoned mathematician knows that simplification is the ultimate shortcut.

Let $\theta ={\sin }^{-1}x$. This implies $\sin \theta =x$. We know from the fundamental identity of trigonometry that ${\cos }^2\theta =1-{\sin }^2\theta$.

Substituting our value for $\sin \theta$, we get ${\cos }^2({\sin }^{-1}x)=1-x^2$. Suddenly, our curve equation transforms into something much more manageable:

By clearing the trigonometric clutter, we have already won half the battle.

To define a line, we need two things: a point and a slope. We are given $x=0$.

Let us find the corresponding $y$-coordinate. Substituting $x=0$ into our simplified equation, we get $y=(1+0{)}^{2y}+1-0^2$, which simplifies to $y=1^{2y}+1$.

Since $1$ raised to any finite power is $1$, we find $y=1+1=2$. Our point of tangency is $P(0,2)$. We have our anchor.

Now, we need the slope. We must differentiate $y=(1+x{)}^{2y}+1-x^2$ with respect to $x$.

The term $(1+x{)}^{2y}$ is the heart of the challenge. Using the derivative rule for $u^v$, where $u=1+x$ and $v=2y$, we apply:

Applying this to our equation, we get:

This looks complex, but watch what happens when we substitute our point $(0,2)$. At $x=0$ and $y=2$, the term $\ln (1+0)$ becomes $\ln (1)$, which is $0$.

The entire term involving the derivative on the right side vanishes! We are left with:

The slope of the tangent $m_t$ is $4$.

We are almost there. The normal is the line perpendicular to the tangent. If the tangent slope is $m_t=4$, the normal slope $m_n$ must satisfy $m_t\cdot m_n=-1$.

Thus, $m_n=-\frac{1}{4}$. Using the point-slope form $y-y_1=m_n(x-x_1)$, we substitute our point $(0,2)$ and our slope $-\frac{1}{4}$:

Multiplying by $4$ gives $4y-8=-x$, or $x+4y=8$.

Look at that. Through careful simplification and systematic differentiation, we have tamed the equation. The beauty of mathematics lies in this: no matter how chaotic an expression appears, there is always a path to order.