If y = y(x) and it follows the relation x cos y + y cos x = π then y''(0) =
(a) 1
(b) -1
(c) π - 1
(

Q: Wait, why do we need to find $y(0)$ first? Can't we just differentiate directly?

Great question! While you could technically differentiate first, you would be left with an expression for $y''$ that still contains $y$ and $y'$. To find the numerical value of $y''(0)$, you *must* know the values of $y$ and $y'$ at $x=0$. Finding $y(0)$ is the essential first step to 'unlock' the values needed for the final substitution.

Q: I keep getting confused with the Product Rule and Chain Rule combined. Any tips?

It is completely normal to feel that way! The trick is to be methodical. Treat each term as a separate mini-problem. For $x \cos y$, write it out: (derivative of $x$) $\cdot \cos y + x \cdot$ (derivative of $\cos y$). The derivative of $\cos y$ is $-\sin y \cdot y'$. If you write it out step-by-step like this, you won't miss the $y'$ term.

Q: Why do so many terms disappear when we substitute $x=0$?

That is the beauty of this problem! Because we are evaluating at $x=0$, any term multiplied by $x$ becomes zero. Furthermore, since $y(0) = \pi$, any term involving $\sin(y)$ becomes $\sin(\pi)$, which is also zero. It's not a coincidence; the problem is designed to reward you for your careful algebraic work by simplifying significantly at the end.

Question

If $y = y(x)$ and it follows the relation $x \cos y + y \cos x = \pi$ then $y''(0) = $

Options:

(A) 1

(B) -1

(C) $\pi - 1$

(D) -\pi

Sigma Percentile · Accepted Answer

### The Dance of Implicit Differentiation

Welcome, future engineers! Today, we are going to peel back the layers of a classic JEE Advanced calculus problem. We are given the implicit relation $x \cos y + y \cos x = \pi$, and our mission is to find the second derivative, $y''(0)$.

This problem is not just about crunching numbers; it is about understanding the hidden geometry of a curve defined implicitly.

### Phase 1: Finding Our Starting Point

Before we can talk about slopes or curvature, we need to know where we are. We are looking for $y''(0)$, which means we need to evaluate our derivatives at $x=0$.

Let's substitute $x=0$ into our original equation:

$$0 \cdot \cos y + y \cos 0 = \pi$$

Since $0 \cdot \cos y$ vanishes and $\cos 0 = 1$, we are left with $y = \pi$. So, our curve passes through the point $(0, \pi)$.

This is our anchor point. Mark it well, for it will be the key to simplifying our expressions later.

### Phase 2: The First Derivative

Now, we differentiate with respect to $x$. Because $y$ is a function of $x$, we must treat it with care. We apply the Product Rule to both $x \cos y$ and $y \cos x$:

$$\frac{d}{dx}(x \cos y) + \frac{d}{dx}(y \cos x) = \frac{d}{dx}(\pi)$$

Applying the rules, we get:

$$(\cos y - x y' \sin y) + (y' \cos x - y \sin x) = 0$$

Now, let's find the slope $y'(0)$ by plugging in $x=0$ and $y=\pi$:

$$\cos \pi - 0 \cdot y'(0) \sin \pi + y'(0) \cos 0 - \pi \sin 0 = 0$$

Since $\cos \pi = -1$, $\cos 0 = 1$, and $\sin 0 = 0$, the equation simplifies to:

$$-1 + y'(0) = 0 \implies y'(0) = 1$$

We have found the slope of the tangent line at our point! It is $1$. This is a crucial milestone.

### Phase 3: The Second Derivative

This is where most students stumble, but we will be precise. We differentiate our first derivative relation again:

$$\frac{d}{dx} [ \cos y - x y' \sin y + y' \cos x - y \sin x ] = 0$$

This requires a meticulous application of the Product Rule. Let's break it down term by term:

1. $\frac{d}{dx}(\cos y) = -y' \sin y$

2. $\frac{d}{dx}(-x y' \sin y) = -y' \sin y - x y'' \sin y - x (y')^2 \cos y$

3. $\frac{d}{dx}(y' \cos x) = y'' \cos x - y' \sin x$

4. $\frac{d}{dx}(-y \sin x) = -y' \sin x - y \cos x$

Combining these, we get a long, intimidating expression. But do not fear! We only need to evaluate it at $x=0, y=\pi, y'=1$.

Watch as the terms vanish:

$$-y' \sin y - y' \sin y - x y'' \sin y - x (y')^2 \cos y + y'' \cos x - y' \sin x - y' \sin x - y \cos x = 0$$

Substituting $x=0, y=\pi, y'=1$:

$$-1 \sin \pi - 1 \sin \pi - 0 - 0 + y''(0) \cos 0 - 1 \sin 0 - 1 \sin 0 - \pi \cos 0 = 0$$

Since $\sin \pi = 0$ and $\sin 0 = 0$, the equation collapses into:

$$0 - 0 + y''(0) \cdot 1 - \pi = 0$$

Which gives us the elegant result:

**$y''(0) = \pi$**

Animated Solution for Mathematics - Differentiation: If $y = y (x)$ and it follows the relation $x cos y + y cos x = π$ then $y^{''} (0) =$

Select Answer:

Visualized Solution

The Implicit Relation

Finding $y (0)$

Differentiating with Respect to $x$

Applying the Product Rule

Finding the Slope $y^{'} (0)$

Preparing for the Second Derivative

Term-by-Term Differentiation

The Complete Second Derivative Equation

Substituting $x = 0, y = π, y^{'} = 1$

Solving for $y^{''} (0)$

Summary of the Solution

The Sigma Insight: Higher Order Derivatives

The Dance of Implicit Differentiation

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) and it follows the relation xcosy+ycosx=π then y′′(0)=

Select Answer:

Visualized Solution

The Implicit Relation

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(0)

Differentiating with Respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Applying the Product Rule

Finding the Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′(0)

Preparing for the Second Derivative

Term-by-Term Differentiation

The Complete Second Derivative Equation

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0,y=π,y′=1

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y′′(0)