JEE Advanced 2000

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Animated Solution for Mathematics - Differentiation: If $x^{2} + y^{2} = 1$ then

Q: Why can't I just write $\frac{d}{dx}(y^2) = 2y$? Why the $y'$?

This is a classic trap! Remember that $y$ is a function of $x$, not an independent variable. When you differentiate a function of $y$ with respect to $x$, you must apply the Chain Rule. So, $\frac{d}{dx}(y^2) = \frac{d}{dy}(y^2) \cdot \frac{dy}{dx} = 2y \cdot y'$. Always remember that $y$ is hiding a $y'$ whenever you differentiate it!

Q: Is there a faster way to solve this without implicit differentiation?

You could solve for $y$ explicitly as $y = \pm \sqrt{1-x^2}$ and then differentiate, but that path is fraught with messy square roots and potential sign errors. Implicit differentiation is the elegant, standard, and much safer way to handle curves defined by algebraic equations.

Select Answer:

Visualized Solution

Visualizing the Unit Circle

We are given the equation of a curve: $x^{2} + y^{2} = 1$ .
This is the standard equation of a unit circle.
Its center is at the origin $(0, 0)$ and its radius is $1$ .

Point and Tangent Geometry

Let $P (x, y)$ be any point on this circle.
The vector from the origin to $P$ is the radius vector.
At point $P$ , we can draw a tangent line to the circle.

First Differentiation Setup

To find the relation, we perform implicit differentiation with respect to $x$ .
We apply the derivative operator $\frac{d}{d x}$ to both sides:
$\frac{d}{d x} (x^{2} + y^{2}) = \frac{d}{d x} (1)$

Applying the Chain Rule

Derivative of $x^{2}$ is $2 x$ .
For $y^{2}$ , we must use the Chain Rule:
$\frac{d}{d x} (y^{2}) = 2 y \cdot \frac{d y}{d x} = 2 y y^{'}$
The derivative of the constant $1$ is $0$ .

Simplifying to $x + y y^{'} = 0$

Our equation is: $2 x + 2 y y^{'} = 0$
Divide both sides by $2$ to simplify:
$x + y y^{'} = 0$
This is our first-order differential relation.

Second Differentiation Setup

To find a relation containing $y^{''}$ , we differentiate again with respect to $x$ :
$\frac{d}{d x} (x + y y^{'}) = \frac{d}{d x} (0)$
We must apply the Product Rule to the term $y y^{'}$ .

Applying the Product Rule

Derivative of $x$ is $1$ .
For $y y^{'}$ , use the Product Rule: $[uv]^{'} = u^{'} v + u v^{'}$
Here, $u = y$ and $v = y^{'}$ :
$\frac{d}{d x} (y y^{'}) = (y)^{'} \cdot y^{'} + y \cdot (y^{'})^{'}$
$\frac{d}{d x} (y y^{'}) = (y^{'})^{2} + y y^{''}$

Final Assembly and Matching

Substitute the derivatives back into the equation:
$1 + (y^{'})^{2} + y y^{''} = 0$
Rearranging the terms:
$y y^{''} + (y^{'})^{2} + 1 = 0$
This matches Option 2.

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The Sigma Insight: Higher Order Derivatives

Solution Diagram

Analyzing the Setup

The unit circle is defined by the equation:

x^{2} + y^{2} = 1

This equation serves as a constraint, defining a boundary for a particle moving along the curve. To understand the motion of a point

(x, y)

on this circle, we must determine the relationship between its coordinates and their derivatives.

The First Derivative

We begin by applying the operator

\frac{d}{d x}

to both sides of the equation. Since

y

is a function of

x

, we employ implicit differentiation and the chain rule:

\frac{d}{d x} (x^{2} + y^{2}) = \frac{d}{d x} (1)

2 x + 2 y \frac{d y}{d x} = 0

Dividing by

2

, we arrive at our first milestone:

x + y y^{'} = 0

This result reveals that the position vector

(x, y)

and the velocity vector

(1, y^{'})

are perpendicular, as their dot product is zero.

The Second Derivative

To find the second derivative

y^{''}

, we differentiate the expression

x + y y^{'} = 0

with respect to

x

once more:

\frac{d}{d x} (x + y y^{'}) = \frac{d}{d x} (0)

Applying the derivative to each term, we get:

1 + \frac{d}{d x} (y y^{'}) = 0

For the term

y y^{'}

, we must use the product rule, where

(uv)^{'} = u^{'} v + u v^{'}

. Setting

u = y

and

v = y^{'}

, we obtain:

1 + (y^{'})^{2} + y y^{''} = 0

Final Result

The expression above represents the elegant differential equation governing the curvature of the unit circle:

y y^{''} + (y^{'})^{2} + 1 = 0

By following these fundamental rules of calculus, we have successfully uncovered the hidden law of the circle's geometry. Keep this logical framework in your toolkit, as it is essential for solving more complex problems in differential geometry.

Common Traps & FAQs

Why do we differentiate twice? Can't we just stop at the first derivative?

Great question! Look at the options provided in the problem. They all contain

y^{''}

, which is the second derivative. This is a hint that the differential equation we are looking for must involve the second derivative. Therefore, we must differentiate our first-order relation

x + y y^{'} = 0

once more.

Why can't I just write

\frac{d}{d x} (y^{2}) = 2 y

? Why the

y^{'}

This is a classic trap! Remember that

y

is a function of

x

, not an independent variable. When you differentiate a function of

y

with respect to

x

, you must apply the Chain Rule. So,

\frac{d}{d x} (y^{2}) = \frac{d}{d y} (y^{2}) \cdot \frac{d y}{d x} = 2 y \cdot y^{'}

. Always remember that

y

is hiding a

y^{'}

whenever you differentiate it!

Is there a faster way to solve this without implicit differentiation?

You could solve for

y

explicitly as

y = \pm 1 - x^{2}

and then differentiate, but that path is fraught with messy square roots and potential sign errors. Implicit differentiation is the elegant, standard, and much safer way to handle curves defined by algebraic equations.

Animated Solution for Mathematics - Differentiation: If $x^{2} + y^{2} = 1$ then

Select Answer:

Visualized Solution

Visualizing the Unit Circle

Point and Tangent Geometry

First Differentiation Setup

Applying the Chain Rule

Simplifying to $x + y y^{'} = 0$

Second Differentiation Setup

Applying the Product Rule

Final Assembly and Matching

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The First Derivative

The Second Derivative

Final Result

Similar Questions

If $x^{2} + y^{2} = 1$ , then

If $cos^{- 1} (\frac{y}{2}) = lo g_{e} (\frac{x}{5})^{5}, ∣ y ∣ < 2$ , then :

If $y = tan^{- 1} (sec x^{3} - tan x^{3})$ , $\frac{π}{2} < x^{3} < \frac{3 π}{2}$ , then

If $y = y (x)$ and it follows the relation $x cos y + y cos x = π$ then $y^{''} (0) =$

If $y^{2} + lo g_{e} (cos^{2} x) = y, x \in (- \frac{π}{2}, \frac{π}{2})$ , then :

$\frac{d ^{2} x}{d y ^{2}}$ equals

If $y = (x + 1 + x^{2})^{n}$ , then $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}$ is

If $y (θ) = \frac{2 c o s θ + c o s 2 θ}{c o s 3 θ + 4 c o s 2 θ + 5 c o s θ + 2}$ , then at $θ = \frac{π}{2}$ , $y^{''} + y^{'} + y$ is equal to :

$\frac{d ^{2} x}{d y ^{2}}$ equals

If $lo g_{e} y = 3 sin^{- 1} x$ , then $(1 - x^{2}) y^{''} - x y^{'}$ at $x = \frac{1}{2}$ is equal to

Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If x2+y2=1 then

Select Answer:

Visualized Solution

Visualizing the Unit Circle

Point and Tangent Geometry

First Differentiation Setup

Applying the Chain Rule

Simplifying to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+yy′=0

Second Differentiation Setup

Applying the Product Rule

Final Assembly and Matching

The Sigma Insight: Higher Order Derivatives

Analyzing the Setup

The First Derivative

The Second Derivative

Final Result

Similar Questions

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Animated Solution for Mathematics - Differentiation: If $x^{2} + y^{2} = 1$ then

Simplifying to $x + y y^{'} = 0$

If $x^{2} + y^{2} = 1$ , then

If $cos^{- 1} (\frac{y}{2}) = lo g_{e} (\frac{x}{5})^{5}, ∣ y ∣ < 2$ , then :

If $y = tan^{- 1} (sec x^{3} - tan x^{3})$ , $\frac{π}{2} < x^{3} < \frac{3 π}{2}$ , then

If $y = y (x)$ and it follows the relation $x cos y + y cos x = π$ then $y^{''} (0) =$

If $y^{2} + lo g_{e} (cos^{2} x) = y, x \in (- \frac{π}{2}, \frac{π}{2})$ , then :

$\frac{d ^{2} x}{d y ^{2}}$ equals

If $y = (x + 1 + x^{2})^{n}$ , then $(1 + x^{2}) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x}$ is

If $y (θ) = \frac{2 c o s θ + c o s 2 θ}{c o s 3 θ + 4 c o s 2 θ + 5 c o s θ + 2}$ , then at $θ = \frac{π}{2}$ , $y^{''} + y^{'} + y$ is equal to :

$\frac{d ^{2} x}{d y ^{2}}$ equals

If $lo g_{e} y = 3 sin^{- 1} x$ , then $(1 - x^{2}) y^{''} - x y^{'}$ at $x = \frac{1}{2}$ is equal to