JEE Advanced 1993

LEVELJEE Main

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

Visualized Solution

Understanding the Objective

We need to find the equation of the normal to the curve:
$y = (1 + x)^{y} + sin^{- 1} (sin^{2} x)$
The point of interest is at $x = 0$ .

Finding the Point of Contact

Substitute $x = 0$ into the original equation to find the $y$ -coordinate.
$y = (1 + 0)^{y} + sin^{- 1} (sin^{2} 0)$
$y = 1^{y} + sin^{- 1} (0)$

Evaluating the $y$ -coordinate

Since $1^{y} = 1$ for any finite $y$ , and $sin^{- 1} (0) = 0$ :
$y = 1 + 0$
$y = 1$
The point of contact is $P (0, 1)$ .

Differentiation Strategy

To find the normal, we first need the slope of the tangent, $m_{t} = \frac{d y}{d x}$ .
Let $u = (1 + x)^{y}$ and $v = sin^{- 1} (sin^{2} x)$ .
Then $y = u + v$ , so $\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$ .

Differentiating the First Term ( $u$ )

For $u = (1 + x)^{y}$ , we use logarithmic differentiation.
Take natural log on both sides: $ln u = y ln (1 + x)$
Differentiate implicitly with respect to $x$ :
$\frac{1}{u} \frac{d u}{d x} = \frac{d y}{d x} ln (1 + x) + y \cdot \frac{1}{1 + x}$

Evaluating $\frac{d u}{d x}$ at $(0, 1)$

Instead of finding a general expression, substitute $x = 0, y = 1$ , and $u = 1$ immediately.
$\frac{1}{1} \frac{d u}{d x} = y^{'} ln (1 + 0) + 1 \cdot \frac{1}{1 + 0}$
$\frac{d u}{d x} = y^{'} \cdot 0 + 1 = 1$

Differentiating the Second Term ( $v$ )

For $v = sin^{- 1} (sin^{2} x)$ , apply the Chain Rule.
$\frac{d v}{d x} = \frac{1}{1 - ( s i n ^{2} x ) ^{2}} \cdot \frac{d}{d x} (sin^{2} x)$
$\frac{d v}{d x} = \frac{1}{1 - s i n ^{4} x} \cdot (2 sin x cos x)$

Evaluating $\frac{d v}{d x}$ at $x = 0$

Substitute $x = 0$ into the derivative of $v$ :
$\frac{d v}{d x} = \frac{1}{1 - s i n ^{4} ( 0 )} \cdot (2 sin (0) cos (0))$
Since $sin (0) = 0$ , the entire expression becomes $0$ .
$\frac{d v}{d x} = 0$

Slope of the Tangent ( $m_{t}$ )

Combine the derivatives to find $\frac{d y}{d x}$ at $x = 0$ :
$m_{t} = \frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$
$m_{t} = 1 + 0 = 1$
The slope of the tangent line is $1$ .

Slope of the Normal ( $m_{n}$ )

The normal line is perpendicular to the tangent line.
Therefore, $m_{n} \cdot m_{t} = - 1$
$m_{n} \cdot (1) = - 1 \Rightarrow m_{n} = - 1$
The slope of the normal is $- 1$ .

Final Equation of the Normal

Use the point-slope form: $y - y_{1} = m_{n} (x - x_{1})$
Substitute the point $(0, 1)$ and slope $m_{n} = - 1$ :
$y - 1 = - 1 (x - 0)$
$y - 1 = - x$
Rearranging gives the final equation: $x + y = 1$

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The Sigma Insight: Tangents, Normals and Rate Measure

Solution Diagram

Analyzing the Setup

The given curve is defined by the equation:

y = (1 + x)^{y} + sin^{- 1} (sin^{2} x)

While the expression appears complex, we can dismantle it systematically to find the equation of the normal line at

x = 0

Phase 1

Finding the Point of Contact

To find the point of contact

P (x_{1}, y_{1})

, we substitute

x = 0

into the original equation:

y = (1 + 0)^{y} + sin^{- 1} (sin^{2} 0)

Since

1^{y} = 1

for any finite

y

and

sin^{- 1} (0) = 0

, we find

y = 1 + 0 = 1

. Thus, our anchor point is

P (0, 1)

Phase 2

The Divide and Conquer Strategy

We need the slope of the tangent,

m_{t} = \frac{d y}{d x}

. To avoid messy differentiation, we define

y = u + v

, where:

u = (1 + x)^{y} and v = sin^{- 1} (sin^{2} x)

The derivative is then simply:

\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}

Phase 3

Taming the Variable Exponent

For

u = (1 + x)^{y}

, we apply logarithmic differentiation:

ln u = y ln (1 + x)

Differentiating implicitly with respect to

x

gives:

\frac{1}{u} \frac{d u}{d x} = \frac{d y}{d x} ln (1 + x) + y \cdot \frac{1}{1 + x}

Substituting

x = 0, y = 1, u = 1

into this expression, the term

ln (1 + 0)

vanishes, leaving:

\frac{d u}{d x} = 1

Phase 4

The Chain Rule Dance

Next, we differentiate

v = sin^{- 1} (sin^{2} x)

using the chain rule:

\frac{d v}{d x} = \frac{1}{1 - ( sin ^{2} x ) ^{2}} \cdot \frac{d}{d x} (sin^{2} x)

The derivative of

sin^{2} x

2 sin x cos x

. Substituting

x = 0

makes the numerator

2 sin (0) cos (0) = 0

, so:

\frac{d v}{d x} = 0

Phase 5

The Final Assembly

We combine our results to find the slope of the tangent:

m_{t} = \frac{d u}{d x} + \frac{d v}{d x} = 1 + 0 = 1

Since the normal is perpendicular to the tangent, its slope

m_{n}

must satisfy

m_{n} \cdot m_{t} = - 1

, yielding

m_{n} = - 1

Using the point-slope form

y - y_{1} = m_{n} (x - x_{1})

with

P (0, 1)

and

m_{n} = - 1

y - 1 = - 1 (x - 0)

The final equation of the normal line is $x + y = 1$ .

Common Traps & FAQs

Why can't we just differentiate the whole expression at once without splitting it?

You technically could, but it would be a nightmare of algebraic bookkeeping. By splitting the function into

u

and

v

, you isolate the logarithmic differentiation (needed for the variable exponent) from the chain rule (needed for the inverse trig). This 'Divide and Conquer' approach minimizes the chance of a sign error or a missing term.

Is it always safe to use logarithmic differentiation when we see a variable in the exponent?

Yes, absolutely! Whenever you see a variable in the base and a variable in the exponent, like

(1 + x)^{y}

, logarithmic differentiation is your most powerful tool. It transforms the exponent into a multiplier, making the differentiation process straightforward.

Why is the slope of the normal the negative reciprocal of the tangent slope?

This is a fundamental geometric property. A normal line is, by definition, perpendicular to the tangent line at the point of contact. If two lines are perpendicular, the product of their slopes must be

- 1

. Therefore, if the tangent slope is

m_{t}

, the normal slope must be

m_{n} = - \frac{1}{m _{t}}

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

Visualized Solution

Understanding the Objective

Finding the Point of Contact

Evaluating the $y$ -coordinate

Differentiation Strategy

Differentiating the First Term ( $u$ )

Evaluating $\frac{d u}{d x}$ at $(0, 1)$

Differentiating the Second Term ( $v$ )

Evaluating $\frac{d v}{d x}$ at $x = 0$

Slope of the Tangent ( $m_{t}$ )

Slope of the Normal ( $m_{n}$ )

Final Equation of the Normal

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

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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; } Find the equation of the normal to the curve y=(1+x)y+sin−1(sin2x) at x=0

Visualized Solution

Understanding the Objective

Finding the Point of Contact

Evaluating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate

Differentiation Strategy

Differentiating the First Term ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdu​ at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,1)

Differentiating the Second Term ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v)

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dxdv​ at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Slope of the Tangent ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mt​)

Slope of the Normal ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mn​)

Final Equation of the Normal

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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

Evaluating the $y$ -coordinate

Differentiating the First Term ( $u$ )

Evaluating $\frac{d u}{d x}$ at $(0, 1)$

Differentiating the Second Term ( $v$ )

Evaluating $\frac{d v}{d x}$ at $x = 0$

Slope of the Tangent ( $m_{t}$ )

Slope of the Normal ( $m_{n}$ )

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

Consider: $f (x) = tan^{- 1} (\frac{1 + s i n x}{1 - s i n x}), x \in (0, \frac{π}{2})$ . A normal to $y = f (x)$ at $x = \frac{π}{6}$ also passes through the point

If the tangent to the curve $y = x + sin y$ at a point $(a, b)$ is parallel to the line joining $(0, \frac{3}{2})$ and $(\frac{1}{2}, 2)$ , then:

The normal to the curve $x = a (1 + cos θ)$ , $y = a sin θ$ at ' $θ$ ' always passes through the fixed point

The normal to the curve $y (x - 2) (x - 3) = x + 6$ at the point where the curve intersects the y-axis passes through the point:

The normal to the curve $x = a (cos θ + θ sin θ)$ , $y = a (sin θ - θ cos θ)$ at any point ' $θ$ ' is such that

The normal to the curve $x = a (cos θ + θ sin θ)$ , $y = a (sin θ - θ cos θ)$ at any point ' $θ$ ' is such that

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Let $P (h, k)$ be a point on the curve $y = x^{2} + 7 x + 2$ , nearest to the line, $y = 3 x - 3$ . Then the equation of the normal to the curve at $P$ is: