Step 1: Simplifying the Inverse Trigonometric Function

Given the equation: $y = \sec^{-1}\left(\frac{1}{2x^2-1}\right)$.

Utilize the trigonometric identity for the reciprocal of the secant function: $\sec^{-1}\left(\frac{1}{u}\right) = \cos^{-1}(u)$ [Per the property of inverse trigonometric functions, where $|u| \geq 1$].

Applying this to the expression:

$y = \cos^{-1}(2x^2 - 1)$

Step 2: Trigonometric Substitution

Given the domain constraint $0 < x < \frac{1}{\sqrt{2}}$, we choose a substitution that simplifies the expression $2x^2 - 1$.

Let $x = \cos \theta$. Then $\theta = \cos^{-1}x$.

Substituting into the expression:

$y = \cos^{-1}(2\cos^2\theta - 1)$

Applying the double-angle formula for cosine: $\cos(2\theta) = 2\cos^2\theta - 1$ [By trigonometric identity]:

$y = \cos^{-1}(\cos 2\theta)$

Step 3: Evaluating the Range and Simplifying

Since $0 < x < \frac{1}{\sqrt{2}}$, it follows that:

$0 < \cos \theta < \frac{1}{\sqrt{2}}$

This implies $\frac{\pi}{4} < \theta < \frac{\pi}{2}$.

Multiplying by 2: $\frac{\pi}{2} < 2\theta < \pi$.

Since $2\theta \in (\frac{\pi}{2}, \pi)$, the expression $\cos^{-1}(\cos 2\theta)$ simplifies to $2\theta$ [By the property $\cos^{-1}(\cos \alpha) = \alpha$ for $\alpha \in [0, \pi]$].

Substituting $\theta = \cos^{-1}x$ back into the equation:

$y = 2\cos^{-1}x$

Step 4: Differentiation

Now, differentiate with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} [2\cos^{-1}x]$

Applying the constant multiple rule and the derivative formula $\frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}$:

$\frac{dy}{dx} = 2 \cdot \left( -\frac{1}{\sqrt{1-x^2}} \right)$

Step 5: Final Result

The derivative is computed as follows:

$\frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}}$

Final Solution: \frac{dy}{dx} = -\frac{2}{\sqrt{1-x^2}}

Given Equation & Differentiation Strategy

The given equation is:

$ax + by^2 = \cos y$

To find $\frac{dy}{dx}$, we must perform implicit differentiation with respect to $x$ across the entire equation [Per the Chain Rule of Calculus].

Step 1: Differentiating Both Sides with Respect to $x$

Applying the operator $\frac{d}{dx}$ to both sides of the equation:

$\frac{d}{dx}(ax + by^2) = \frac{d}{dx}(\cos y)$

Step 2: Applying Sum and Constant Multiple Rules

Distributing the derivative across the terms and factoring out constants:

$a \cdot \frac{d}{dx}(x) + b \cdot \frac{d}{dx}(y^2) = \frac{d}{dx}(\cos y)$

[By the Linearity Property of Differentiation]

Step 3: Executing the Chain Rule

Differentiate each term individually, keeping in mind that $y$ is a function of $x$ ($y=f(x)$):

• $\frac{d}{dx}(ax) = a(1) = a$
• $b \cdot \frac{d}{dx}(y^2) = b \cdot (2y \cdot \frac{dy}{dx}) = 2by \frac{dy}{dx}$ [By Power Rule and Chain Rule]
• $\frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx}$ [By Derivative of Trigonometric Functions and Chain Rule]

Substituting these back into the equation:

$a + 2by \frac{dy}{dx} = -\sin y \frac{dy}{dx}$

Step 4: Isolating the $\frac{dy}{dx}$ terms

Group all terms containing $\frac{dy}{dx}$ on one side of the equation:

$a = -\sin y \frac{dy}{dx} - 2by \frac{dy}{dx}$

$a = -\frac{dy}{dx} (\sin y + 2by)$

Step 5: Solving for $\frac{dy}{dx}$

Divide both sides by the coefficient of $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{a}{ -(\sin y + 2by)}$

$\frac{dy}{dx} = -\frac{a}{2by + \sin y}$

Geometric Representation of Implicit Differentiation

Final Solution: $\frac{dy}{dx} = -\frac{a}{2by + \sin y}$

Step 1: Analyzing the Trigonometric Substitution

Given the expression $y = \sin^{-1} \left( \frac{2x}{1+x^2} \right)$, we recognize that the interior fraction resembles the double-angle identity for sine: $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$.

To simplify the derivative, we perform the substitution $x = \tan\theta$, which implies $\theta = \tan^{-1}(x)$.

[Justification: Per trigonometric identity $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$].

Step 2: Transforming the Function

Substituting $x = \tan\theta$ into the equation:

$y = \sin^{-1} \left( \frac{2\tan\theta}{1+\tan^2\theta} \right)$
$y = \sin^{-1} (\sin 2\theta)$
$y = 2\theta$

[Justification: Since $\sin^{-1}(\sin \alpha) = \alpha$ for $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$].

Step 3: Re-substituting $x$ and Differentiating

Replacing $\theta$ with $\tan^{-1}(x)$:

$y = 2\tan^{-1}(x)$

Now, we differentiate both sides with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} [2\tan^{-1}(x)]$
$\frac{dy}{dx} = 2 \cdot \frac{d}{dx} [\tan^{-1}(x)]$

[Justification: By the Constant Multiple Rule of differentiation].

Step 4: Applying Standard Derivative Formula

Using the standard derivative formula for inverse trigonometric functions:

$\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1+x^2}$
$\frac{dy}{dx} = 2 \cdot \left( \frac{1}{1+x^2} \right)$

[Justification: Differentiation of inverse tangent function].

Step 5: Visual Representation of the Differentiation Geometry

Derivative Step	Result
Substitution	$x = \tan\theta$
Simplified Function	$y = 2\tan^{-1}(x)$
Resulting Derivative	$\frac{2}{1+x^2}$

Final Solution: \frac{dy}{dx} = \frac{2}{1+x^2}

Step 1: Application of Differentiation Operator

Given the implicit function: $2x + 3y = \sin y$. To find $\frac{dy}{dx}$, we apply the derivative operator $\frac{d}{dx}$ to both sides of the equation with respect to $x$:

$\frac{d}{dx}(2x + 3y) = \frac{d}{dx}(\sin y)$

Step 2: Linear Property and Chain Rule Application

By the linearity of the derivative, we distribute the operator. Furthermore, we apply the Chain Rule [$\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$] to the terms containing $y$:

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$

$2 \cdot \frac{d}{dx}(x) + 3 \cdot \frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$ [Per the differentiation rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the chain rule for trigonometric functions]

Step 3: Algebraic Simplification

Substituting $\frac{d}{dx}(x) = 1$, the equation becomes:

$2 + 3\frac{dy}{dx} = \cos y \frac{dy}{dx}$

Step 4: Isolation of the Derivative term

To solve for $\frac{dy}{dx}$, we group all terms containing $\frac{dy}{dx}$ on one side of the equation:

$2 = \cos y \frac{dy}{dx} - 3\frac{dy}{dx}$

$2 = \frac{dy}{dx}(\cos y - 3)$ [By distributive property of multiplication]

Step 5: Final Evaluation

Divide both sides by the coefficient of $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{2}{\cos y - 3}$

Final Solution: $\frac{dy}{dx} = \frac{2}{\cos y - 3}$

Step 1: Application of Implicit Differentiation

Given the equation: $\sin^2 y + \cos(xy) = \kappa$, where $\kappa$ is a constant. We differentiate both sides with respect to $x$ using the chain rule and the sum rule [Per the Rule of Differentiation of Implicit Functions].

$\frac{d}{dx}(\sin^2 y) + \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(\kappa)$

Step 2: Differentiating Term by Term

For the first term $\frac{d}{dx}(\sin^2 y)$: Applying the power rule followed by the chain rule: $\frac{d}{dx}(\sin^2 y) = 2 \sin y \cdot \frac{d}{dx}(\sin y) = 2 \sin y \cdot \cos y \cdot \frac{dy}{dx}$ Using the double angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$, we simplify: $= \sin(2y) \frac{dy}{dx}$

For the second term $\frac{d}{dx}(\cos(xy))$: Applying the chain rule and product rule: $\frac{d}{dx}(\cos(xy)) = -\sin(xy) \cdot \frac{d}{dx}(xy)$ Using the product rule for $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y(1)$: $= -\sin(xy) \cdot (x \frac{dy}{dx} + y)$

For the constant term: $\frac{d}{dx}(\kappa) = 0$ [Per the Constant Rule of Differentiation]

Step 3: Constructing the Linear Equation for $\frac{dy}{dx}$

Substituting the derivatives back into the original expression: $\sin(2y) \frac{dy}{dx} - \sin(xy)(x \frac{dy}{dx} + y) = 0$

Distributing $-\sin(xy)$: $\sin(2y) \frac{dy}{dx} - x \sin(xy) \frac{dy}{dx} - y \sin(xy) = 0$

Step 4: Isolating $\frac{dy}{dx}$

Group all terms containing $\frac{dy}{dx}$ on one side: $\frac{dy}{dx} (\sin(2y) - x \sin(xy)) = y \sin(xy)$

Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}$

Visual Representation of the Differential Geometry

Final Solution: \frac{dy}{dx} = \frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}

Step 1: Given Equation and Differentiation Strategy

The given implicit function is:

$\sin^2 x + \cos^2 y = 1$

To find $\frac{dy}{dx}$, we must apply the operator $\frac{d}{dx}$ to both sides of the equation. We utilize the Chain Rule, which states that for a composite function $f(g(x))$, $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$ [Per the Chain Rule of Differentiation].

Step 2: Applying the Derivative Operator

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^2 y) = \frac{d}{dx}(1)$

Step 3: Calculating Individual Derivatives

• For the first term, treat $\sin^2 x$ as $(\sin x)^2$: $\frac{d}{dx}(\sin^2 x) = 2 \sin x \cdot \frac{d}{dx}(\sin x) = 2 \sin x \cos x$ [Using the Power Rule and $\frac{d}{dx}\sin x = \cos x$]
• For the second term, treat $\cos^2 y$ as $(\cos y)^2$: $\frac{d}{dx}(\cos^2 y) = 2 \cos y \cdot \frac{d}{dx}(\cos y) = 2 \cos y \cdot (-\sin y) \cdot \frac{dy}{dx}$ [Using the Chain Rule and $\frac{d}{dx}\cos y = -\sin y \frac{dy}{dx}$]
• The derivative of a constant is zero: $\frac{d}{dx}(1) = 0$.

Step 4: Algebraic Rearrangement

Substitute these results back into the equation:

$2 \sin x \cos x - 2 \cos y \sin y \cdot \frac{dy}{dx} = 0$

Apply the double angle identity $\sin(2\theta) = 2 \sin \theta \cos \theta$ [Per Trigonometric Identities]:

$\sin(2x) - \sin(2y) \cdot \frac{dy}{dx} = 0$

Step 5: Solving for $\frac{dy}{dx}$

Isolate the term containing $\frac{dy}{dx}$:

$\sin(2y) \cdot \frac{dy}{dx} = \sin(2x)$

Divide both sides by $\sin(2y)$:

$\frac{dy}{dx} = \frac{\sin(2x)}{\sin(2y)}$

Step 6: Visual Representation

Final Solution: \frac{dy}{dx} = \frac{\sin(2x)}{\sin(2y)}

Given Variables & Initial Setup

The given implicit equation is:

$x^3 + x^2y + xy^2 + y^3 = 81$

To find $\frac{dy}{dx}$, we must apply the process of implicit differentiation with respect to $x$, treating $y$ as a function of $x$ [Per the Chain Rule: $\frac{d}{dx}(f(y)) = f'(y) \cdot \frac{dy}{dx}$].

Step 1: Differentiating Both Sides with Respect to $x$

Applying the derivative operator $\frac{d}{dx}$ to the entire equation:

$\frac{d}{dx}(x^3) + \frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) + \frac{d}{dx}(y^3) = \frac{d}{dx}(81)$

Step 2: Executing Individual Derivatives

Using the Power Rule ($\frac{d}{dx}x^n = nx^{n-1}$) and the Product Rule ($\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$):

• $\frac{d}{dx}(x^3) = 3x^2$
• $\frac{d}{dx}(x^2y) = x^2 \frac{dy}{dx} + y(2x) = x^2 \frac{dy}{dx} + 2xy$
• $\frac{d}{dx}(xy^2) = x(2y \frac{dy}{dx}) + y^2(1) = 2xy \frac{dy}{dx} + y^2$
• $\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}$
• $\frac{d}{dx}(81) = 0$ [Per the Constant Derivative Rule]

Step 3: Assembling the Differential Equation

Substituting the derivatives back into the expression:

$3x^2 + x^2 \frac{dy}{dx} + 2xy + 2xy \frac{dy}{dx} + y^2 + 3y^2 \frac{dy}{dx} = 0$

Step 4: Isolating $\frac{dy}{dx}$

Group terms containing $\frac{dy}{dx}$ on one side and move constant/x-y terms to the other:

$(x^2 + 2xy + 3y^2) \frac{dy}{dx} = -(3x^2 + 2xy + y^2)$

Divide by the coefficient of $\frac{dy}{dx}$:

$\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}$

Visual Representation of the Derivative Slope Field

Final Solution: $\frac{dy}{dx} = -\frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}$

Step 1: Trigonometric Substitution

Given the function $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, we utilize the trigonometric identity for $\cos(2\theta)$ in terms of $\tan\theta$. Let $x = \tan\theta$, where $\theta = \tan^{-1}x$.

Since $0 < x < 1$, we have $0 < \tan\theta < 1$, implying $0 < \theta < \frac{\pi}{4}$ [Domain constraints of inverse trigonometric functions].

Step 2: Simplification of the Expression

Substituting $x = \tan\theta$ into the argument of the function:

$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta}$

Using the identity $\cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta}$ [Trigonometric Double Angle Identity]:

The function becomes $y = \sin^{-1}(\cos 2\theta)$.

Step 3: Conversion to Inverse Sine Domain

To evaluate $\sin^{-1}(\cos 2\theta)$, we convert $\cos(2\theta)$ to sine using the co-function identity $\cos(A) = \sin\left(\frac{\pi}{2} - A\right)$:

$y = \sin^{-1}\left[\sin\left(\frac{\pi}{2} - 2\theta\right)\right]$

Given $0 < \theta < \frac{\pi}{4}$, then $0 < 2\theta < \frac{\pi}{2}$. Thus, $-\frac{\pi}{2} < -2\theta < 0$, and consequently $0 < \frac{\pi}{2} - 2\theta < \frac{\pi}{2}$. This range falls within the principal value branch of $\sin^{-1}(x)$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Therefore, $y = \frac{\pi}{2} - 2\theta$ [By the property $\sin^{-1}(\sin \alpha) = \alpha$ for $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$].

Step 4: Differentiation with respect to x

Substitute back $\theta = \tan^{-1}x$:

$y = \frac{\pi}{2} - 2\tan^{-1}x$

Differentiating both sides with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) - 2\frac{d}{dx}(\tan^{-1}x)$

Since the derivative of a constant is $0$ and $\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}$ [Standard Derivative Formula]:

$\frac{dy}{dx} = 0 - 2\left(\frac{1}{1+x^2}\right)$

Visual Representation of Substitution

Final Solution: $\frac{dy}{dx} = -\frac{2}{1+x^2}$

Step 1: Substitution and Domain Analysis

Given the function $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right)$, we observe the structure of the internal algebraic expression $\frac{2x}{1+x^2}$. This form suggests the trigonometric identity for $\sin(2\theta)$ when we substitute $x = \tan\theta$.

Let $x = \tan\theta$. Given $-1 < x < 1$, the corresponding range for $\theta$ is $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.

[By trigonometric substitution: $\sin(2\theta) = \frac{2\tan\theta}{1+\tan^2\theta}$]

Step 2: Simplifying the Expression

Substitute $x = \tan\theta$ into the function:

$y = \cos^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right)$ $y = \cos^{-1}(\sin(2\theta))$

To evaluate $\cos^{-1}(\sin(2\theta))$, we convert sine to cosine using the co-function identity $\sin(\alpha) = \cos\left(\frac{\pi}{2} - \alpha\right)$:

$y = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - 2\theta\right)\right)$

Since $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, then $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$. Multiplying by $-1$, we get $-\frac{\pi}{2} < -2\theta < \frac{\pi}{2}$. Adding $\frac{\pi}{2}$, we obtain $0 < \frac{\pi}{2} - 2\theta < \pi$.

[Per the definition of the principal value branch of $\cos^{-1}(x)$, which is $[0, \pi]$, we can simplify the expression directly.]

$y = \frac{\pi}{2} - 2\theta$

Step 3: Back-Substitution and Differentiation

Since $x = \tan\theta$, it follows that $\theta = \tan^{-1}(x)$. Substituting this back into the equation for $y$:

$y = \frac{\pi}{2} - 2\tan^{-1}(x)$

Now, differentiate with respect to $x$ using the linearity property of the derivative:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) - 2 \frac{d}{dx}\left(\tan^{-1}(x)\right)$

[Per the Constant Rule, $\frac{d}{dx}(C) = 0$, and per the standard derivative rule, $\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}$]

$\frac{dy}{dx} = 0 - 2\left(\frac{1}{1+x^2}\right)$ $\frac{dy}{dx} = -\frac{2}{1+x^2}$

Visual Representation of the Function Gradient

Final Solution: $\frac{dy}{dx} = -\frac{2}{1+x^2}$

Step 1: Trigonometric Substitution

Given the expression $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, we recognize the identity for $\cos(2\theta)$. We substitute $x = \tan\theta$, which implies $\theta = \tan^{-1}x$.

Since $0 < x < 1$, it follows that $0 < \tan\theta < 1$, which implies $0 < \theta < \frac{\pi}{4}$ [By domain restriction of the inverse tangent function].

Step 2: Simplifying the Inverse Trigonometric Function

Substitute $x = \tan\theta$ into the expression:

$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos(2\theta)$ [Per the double-angle identity for cosine].

Thus, $y = \cos^{-1}(\cos(2\theta))$.

Since $0 < \theta < \frac{\pi}{4}$, then $0 < 2\theta < \frac{\pi}{2}$. Because $2\theta \in [0, \pi]$, we have $\cos^{-1}(\cos(2\theta)) = 2\theta$ [By definition of the principal value branch of $\cos^{-1}$].

Substituting $\theta = \tan^{-1}x$ back into the equation, we obtain:

$y = 2\tan^{-1}x$

Step 3: Differentiation with Respect to $x$

Differentiating both sides with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} [2\tan^{-1}x]$

Applying the constant multiple rule and the standard derivative formula for inverse tangent:

$\frac{d}{dx} [\tan^{-1}x] = \frac{1}{1+x^2}$ [Standard derivative axiom].

$\frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2}$

Visual Representation of the Function Relationship

Final Solution: $\frac{dy}{dx} = \frac{2}{1+x^2}$

Step 1: Identifying the Mathematical Structure

The given equation is an implicit function of the form $f(x, y) = c$, where $2x + 3y = \sin x$. To find the derivative $\frac{dy}{dx}$, we must differentiate both sides of the equation with respect to $x$ [Per the Sum Rule and Chain Rule of Calculus].

Step 2: Differentiating Both Sides with Respect to $x$

We apply the operator $\frac{d}{dx}$ to each term independently:

$$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x)$$

Step 3: Applying Derivative Rules

• For the first term, apply the Power Rule: $\frac{d}{dx}(ax^n) = anx^{n-1}$. Thus, $\frac{d}{dx}(2x) = 2(1) = 2$.
• For the second term, apply the constant multiple rule and the implicit differentiation rule: $\frac{d}{dx}(3y) = 3 \cdot \frac{dy}{dx}$.
• For the right-hand side, apply the standard trigonometric derivative: $\frac{d}{dx}(\sin x) = \cos x$.

Step 4: Algebraic Manipulation to Isolate $\frac{dy}{dx}$

Substituting the results from Step 3 into our equation:

$$2 + 3\frac{dy}{dx} = \cos x$$

Subtracting $2$ from both sides [Per the Subtraction Property of Equality]:

$$3\frac{dy}{dx} = \cos x - 2$$

Dividing by $3$ to isolate the derivative:

$$\frac{dy}{dx} = \frac{\cos x - 2}{3}$$

Visualizing the Gradient Function

Final Solution: $\frac{dy}{dx} = \frac{\cos x - 2}{3}$

Step 1: Establishing the Trigonometric Substitution

Given the expression $y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)$, we observe the internal structure of the argument. It mirrors the standard trigonometric identity for the triple angle of tangent. [By Trigonometric Identity: $\tan(3\theta) = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$].

Let $x = \tan\theta$. Then $\theta = \tan^{-1}(x)$.

Step 2: Determining the Domain Restriction

The given domain is $-\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$. Substituting $x = \tan\theta$: $-\frac{1}{\sqrt{3}} < \tan\theta < \frac{1}{\sqrt{3}}$ $\tan\left(-\frac{\pi}{6}\right) < \tan\theta < \tan\left(\frac{\pi}{6}\right)$ This implies $-\frac{\pi}{6} < \theta < \frac{\pi}{6}$. Multiplying by 3, we get: $-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$. [Justification: Since $-\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$, the expression $3\theta$ lies within the principal branch of the $\tan^{-1}$ function].

Step 3: Simplification of the Function

Substitute $x = \tan\theta$ into the original equation:

$y = \tan^{-1}\left(\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\right)$

[By triple angle identity]:

$y = \tan^{-1}(\tan 3\theta)$

Since $3\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, we have $y = 3\theta$.

Substituting $\theta = \tan^{-1}(x)$ back into the equation:

$y = 3\tan^{-1}(x)$

Step 4: Differentiation with Respect to $x$

Differentiating both sides with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx} [3\tan^{-1}(x)]$

Using the constant multiple rule and the standard derivative of the inverse tangent function [$\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}$]:

$\frac{dy}{dx} = 3 \cdot \left(\frac{1}{1 + x^2}\right)$

Graphical Representation of the Function $y = 3\tan^{-1}(x)$

Final Solution: $\frac{dy}{dx} = \frac{3}{1 + x^2}$

Step 1: Differentiation with respect to $x$

To find $\frac{dy}{dx}$ for the implicit function $xy + y^2 = \tan x + y$, we differentiate both sides of the equation with respect to $x$ using the rules of differentiation [Per the Sum Rule and Chain Rule of Calculus]:

$\frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(\tan x) + \frac{d}{dx}(y)$

Step 2: Applying Product and Chain Rules

We apply the Product Rule, $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$, to the term $xy$, and the Chain Rule to $y^2$:

• $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y(1)$
• $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$ [Per the Chain Rule]
• $\frac{d}{dx}(\tan x) = \sec^2 x$ [Standard derivative formula]
• $\frac{d}{dx}(y) = \frac{dy}{dx}$

Substituting these into our differentiated equation:

$x\frac{dy}{dx} + y + 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx}$

Step 3: Isolating the $\frac{dy}{dx}$ terms

Rearrange the terms to gather all components containing $\frac{dy}{dx}$ on the left-hand side of the equation:

$x\frac{dy}{dx} + 2y\frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x - y$

Step 4: Factoring and Solving for $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ from the left side:

$\frac{dy}{dx}(x + 2y - 1) = \sec^2 x - y$

Divide both sides by $(x + 2y - 1)$ to isolate the derivative [Assuming $x + 2y - 1 \neq 0$]:

$\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}$

Visual Representation of Implicit Differentiation

Final Solution: $\frac{dy}{dx} = \frac{\sec^2 x - y}{x + 2y - 1}$

Step 1: Trigonometric Substitution

To simplify the expression $y = \sin^{-1}(2x\sqrt{1-x^2})$, we use the trigonometric substitution method. Let $x = \sin\theta$. Consequently, $\theta = \sin^{-1}x$. [By Trigonometric Identity: $\sin^2\theta + \cos^2\theta = 1$].

Given the domain $-\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$, we have $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$.

Step 2: Algebraic Simplification

Substitute $x = \sin\theta$ into the function:

$y = \sin^{-1}(2\sin\theta \sqrt{1-\sin^2\theta})$
$y = \sin^{-1}(2\sin\theta \sqrt{\cos^2\theta})$
$y = \sin^{-1}(2\sin\theta \cos\theta)$

Utilizing the double-angle formula $\sin(2\theta) = 2\sin\theta\cos\theta$ [By Double Angle Identity]:

$y = \sin^{-1}(\sin 2\theta)$

Step 3: Determining the Principal Value

Since $-\frac{\pi}{4} < \theta < \frac{\pi}{4}$, it follows that $-\frac{\pi}{2} < 2\theta < \frac{\pi}{2}$.

Because the range of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we can simplify $\sin^{-1}(\sin 2\theta) = 2\theta$. [By definition of inverse trigonometric functions].

$y = 2\theta = 2\sin^{-1}x$

Step 4: Differentiation

Now, differentiate $y$ with respect to $x$ using the standard derivative formula for inverse sine: $\frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$. [By Derivative of Inverse Trigonometric Functions]:

$\frac{dy}{dx} = \frac{d}{dx}(2\sin^{-1}x)$
$\frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\sin^{-1}x)$
$\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$

Visual Representation of Substitution

Final Solution: \frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}

Step 1: Application of the Derivative Operator

To determine $\frac{dy}{dx}$ for the implicit function $x^2 + xy + y^2 = 100$, we apply the differentiation operator $\frac{d}{dx}$ to both sides of the equation with respect to $x$:

$\frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(100)$

[By the Linearity Property of Differentiation: $\frac{d}{dx}(u + v + w) = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx}$]

Step 2: Term-by-Term Differentiation

We differentiate each term individually:

• $\frac{d}{dx}(x^2) = 2x$ [Using the Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$]
• $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y(1)$ [Using the Product Rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$]
• $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ [Using the Chain Rule: $\frac{d}{dx}(f(y)) = \frac{df}{dy} \cdot \frac{dy}{dx}$]
• $\frac{d}{dx}(100) = 0$ [Derivative of a constant is zero]

Step 3: Constructing the Algebraic Equation

Substituting the results from Step 2 into the differentiated equation:

$2x + \left( x\frac{dy}{dx} + y \right) + 2y\frac{dy}{dx} = 0$

Step 4: Isolating the $\frac{dy}{dx}$ Term

Group all terms containing $\frac{dy}{dx}$ on one side and shift remaining terms to the opposite side:

$x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y$

Factor out $\frac{dy}{dx}$ from the left side:

$\frac{dy}{dx}(x + 2y) = -(2x + y)$

Step 5: Solving for the Derivative

Divide both sides by the expression $(x + 2y)$ to solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$

Final Solution: $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$