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Animated Solution for Mathematics - Differentiation: Let $y$ be an implicit function of $x$ defined by $x^{2 x} - 2 x^{x} cot y - 1 = 0$ . Then $y^{'} (1)$ equals

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Progressive Hints

Hint 1

Hint 1 (The Spark): Before diving into derivatives, focus on the point of interest. Substitute

x = 1

into the equation

x^{2 x} - 2 x^{x} cot y - 1 = 0

to find the corresponding value of

y

. You will find that

cot y

must be zero, which tells you the value of

y

Hint 2

Hint 2 (The Arsenal): The term

x^{x}

is a classic source of complexity. Use the substitution

u = x^{x}

to simplify the equation to a quadratic form in terms of

u

. Remember that you will need to find

\frac{d u}{d x}

using logarithmic differentiation, i.e.,

u = e^{x l n x}

Hint 3

Hint 3 (The Execution): After substituting

u = x^{x}

, your equation becomes

u^{2} - 2 u cot y - 1 = 0

. Differentiate this with respect to

x

using the chain rule and product rule. At

x = 1

, you will have

u = 1

u^{'} = 1

, and

cot y = 0

. Use the identity

csc^{2} y = 1 + cot^{2} y

to find

csc^{2} y

and solve for

y^{'} (1)

Visualized Solution

The Implicit Equation

Given equation: $x^{2 x} - 2 x^{x} cot y - 1 = 0$
Goal: Find $y^{'} (1)$
This is an implicit function involving terms like $x^{x}$ and $cot y$ .

Finding $y$ at $x = 1$

Substitute $x = 1$ into the equation:
$1^{2 (1)} - 2 (1)^{1} cot y - 1 = 0$
Simplify: $1 - 2 cot y - 1 = 0$
$- 2 cot y = 0 ⟹ cot y = 0$
Therefore, at $x = 1$ , $y = \frac{π}{2}$

Substitution $u = x^{x}$

Let $u = x^{x}$
The equation becomes: $u^{2} - 2 u cot y - 1 = 0$
At $x = 1$ , $u = 1^{1} = 1$

Derivative of $x^{x}$

Differentiate $u = x^{x}$ with respect to $x$ :
$\frac{d u}{d x} = \frac{d}{d x} (e^{x l n x}) = x^{x} (1 + ln x)$
At $x = 1$ : $u^{'} = 1^{1} (1 + ln 1) = 1 (1 + 0) = 1$

Implicit Differentiation

Differentiate $u^{2} - 2 u cot y - 1 = 0$ with respect to $x$ :
$2 u \frac{d u}{d x} - 2 [\frac{d u}{d x} cot y + u (- csc^{2} y) \frac{d y}{d x}] = 0$
Divide by $2$ : $u u^{'} - u^{'} cot y + u csc^{2} y \cdot y^{'} = 0$

Substituting Known Values

Substitute known values at $x = 1$ :
$u = 1, u^{'} = 1, cot y = 0$
Calculate $csc^{2} y$ : $csc^{2} y = 1 + cot^{2} y = 1 + 0^{2} = 1$
Plug into the differentiated equation:
$(1) (1) - (1) (0) + (1) (1) y^{'} = 0$

Solving for $y^{'} (1)$

Simplify the expression: $1 + y^{'} = 0$
Solve for $y^{'}$ : $y^{'} = - 1$
Final result: $y^{'} (1) = - 1$

00:00 / 00:00

The Sigma Insight: Techniques of Differentiation

Solution Diagram

Analyzing the Setup

The given equation is

x^{2 x} - 2 x^{x} cot y - 1 = 0

. We aim to find the slope of the curve, denoted as

y^{'} (1)

, at the point where

x = 1

Finding the Anchor Point

Before differentiating, we must determine the value of

y

when

x = 1

. Substituting

x = 1

into the equation yields:

1^{2 (1)} - 2 (1)^{1} cot y - 1 = 0

This simplifies to

1 - 2 cot y - 1 = 0

, which further reduces to

- 2 cot y = 0

. Since

cot y = 0

occurs at

y = \frac{π}{2}

, our anchor point is

(1, \frac{π}{2})

The Art of Substitution

To simplify the differentiation, we introduce the helper variable

u = x^{x}

. The original equation transforms into the quadratic form:

u^{2} - 2 u cot y - 1 = 0

We calculate the derivative

u^{'}

using the identity

u = e^{x l n x}

. Applying the chain rule, we find:

u^{'} = x^{x} (1 + ln x)

x = 1

, this evaluates to

u^{'} (1) = 1^{1} (1 + ln 1) = 1 (1 + 0) = 1

. Additionally, at

x = 1

, the value of

u

1^{1} = 1

The Differentiation Dance

We now perform implicit differentiation on the equation

u^{2} - 2 u cot y - 1 = 0

with respect to

x

. Applying the chain rule and the product rule, we obtain:

2 u u^{'} - 2 (u^{'} cot y + u (- csc^{2} y) y^{'}) = 0

Dividing the entire equation by

2

, we isolate the terms:

u u^{'} - u^{'} cot y + u csc^{2} y \cdot y^{'} = 0

Final Calculation

We substitute our known values:

u = 1

u^{'} = 1

, and

cot y = 0

. Since

csc^{2} y = 1 + cot^{2} y

, at our point,

csc^{2} y = 1 + 0^{2} = 1

Substituting these into the differentiated equation gives:

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

This simplifies to

1 + y^{'} = 0

. Therefore, the slope of the tangent line at

x = 1

is:

$y^{'} = - 1$

Common Traps & FAQs

Wait, why do we reject the negative root or why can't we just use energy conservation?

This is a calculus problem, not a physics one, so energy conservation doesn't apply here! Regarding the roots, when we solve

cot y = 0

, we get

y = \frac{π}{2} + nπ

. For the purpose of finding the derivative at a specific point, we can take the principal value

y = \frac{π}{2}

. The derivative will be the same regardless of the integer

n

because the slope of the cotangent function is periodic.

Why can't we just take the natural log of both sides of the original equation?

That is a common trap! You cannot take the natural log of a sum or difference directly (i.e.,

ln (A - B) \neq = ln A - ln B

). Because your equation is

x^{2 x} - 2 x^{x} cot y - 1 = 0

, the subtraction prevents you from using logarithmic properties to simplify the expression. That is exactly why the substitution

u = x^{x}

is the elegant path forward.

How do I handle the derivative of

x^{x}

Great question! Whenever you see a variable raised to a variable power, like

x^{x}

, the standard technique is to write it as an exponential:

x^{x} = e^{l n (x^{x})} = e^{x l n x}

. Then, use the chain rule:

\frac{d}{d x} (e^{x l n x}) = e^{x l n x} \cdot \frac{d}{d x} (x ln x)

. Applying the product rule to

x ln x

gives you

1 \cdot ln x + x \cdot \frac{1}{x} = ln x + 1

. So, the derivative is

x^{x} (1 + ln x)

Animated Solution for Mathematics - Differentiation: Let $y$ be an implicit function of $x$ defined by $x^{2 x} - 2 x^{x} cot y - 1 = 0$ . Then $y^{'} (1)$ equals

Select Answer:

Visualized Solution

The Implicit Equation

Finding $y$ at $x = 1$

Substitution $u = x^{x}$

Derivative of $x^{x}$

Implicit Differentiation

Substituting Known Values

Solving for $y^{'} (1)$

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

Finding the Anchor Point

The Art of Substitution

The Differentiation Dance

Final Calculation

Similar Questions

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If $2 y = (cot^{- 1} (\frac{3 c o s x + s i n x}{c o s x - 3 s i n x}))^{2}, x \in (0, \frac{π}{2})$ , then $\frac{d y}{d x}$ is equal to :

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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; } Let y be an implicit function of x defined by x2x−2xxcoty−1=0. Then y′(1) equals

Select Answer:

Visualized Solution

The Implicit Equation

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

Substitution .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u=xx

Derivative of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xx

Implicit Differentiation

Substituting Known Values

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

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

Finding the Anchor Point

The Art of Substitution

The Differentiation Dance

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Differentiation: Let $y$ be an implicit function of $x$ defined by $x^{2 x} - 2 x^{x} cot y - 1 = 0$ . Then $y^{'} (1)$ equals

Finding $y$ at $x = 1$

Substitution $u = x^{x}$

Derivative of $x^{x}$

Solving for $y^{'} (1)$

If $y$ is a function of $x$ and $lo g (x + y) - 2 x y = 0$ , then the value of $y^{'} (0)$ is equal to

If $2 y = (cot^{- 1} (\frac{3 c o s x + s i n x}{c o s x - 3 s i n x}))^{2}, x \in (0, \frac{π}{2})$ , then $\frac{d y}{d x}$ is equal to :

Find $\frac{d y}{d x}$ at $x = - 1$ , when $(sin y)^{s i n (\frac{π x}{2})} + \frac{3}{2} sec^{- 1} (2 x) + 2^{x} tan (ln (x + 2)) = 0$ .

For $x > 1$ , if $(2 x)^{2 y} = 4 e^{2 x - 2 y}$ , then $(1 + lo g_{e} 2 x)^{2} \frac{d y}{d x}$ is equal to :

If $x^{l o g_{e} (l o g_{e} x)} - x^{2} + y^{2} = 4 (y > 0)$ , then $d y / d x$ at $x = e$ is equal to :

If $y = sec (tan^{- 1} x)$ , then $\frac{d y}{d x}$ at $x = 1$ is equal to

If $y = f (\frac{2 x - 1}{x ^{2} + 1})$ and $f^{'} (x) = sin x^{2}$ , then $\frac{d y}{d x} = \dots\dots\dots$

Let $f (\frac{x + y}{2}) = \frac{f ( x ) + f ( y )}{2}$ for all real $x$ and $y$ . If $f^{'} (0)$ exists and equals $- 1$ and $f (0) = 1$ , find $f (2)$ .

The derivative of $sec^{- 1} (\frac{1}{2 x ^{2} - 1})$ with respect to $1 - x^{2}$ at $x = \frac{1}{2}$ is $\dots\dots\dots$

If the function $f (x) = x^{3} + e^{x /2}$ and $g (x) = f^{- 1} (x)$ , then the value of $g^{'} (1)$ is