2.4 Q-2

Question Statement

Find $\frac{dy}{dx}$ for the following equations:

1. $3x + 4y + 7 = 0$
2. $xy + y^2 = 2$
3. $x^2 - 4xy - 5y = 0$
4. $4x^2 + 2hx y + by^2 + 2gx + 2fy + c = 0$
5. $x \sqrt{1+y} + y \sqrt{1+x} = 0$
6. $y(x^2 - 1) = x \sqrt{x^2 + 4}$

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Background and Explanation

To solve these problems, we use implicit differentiation, which involves differentiating both sides of an equation with respect to $x$, applying the chain rule wherever $y$ appears. This process will yield terms involving $\frac{dy}{dx}$, which can then be isolated to find the derivative.

Key concepts:

• The chain rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$.
• Differentiation of implicit terms like $y$: $\frac{d}{dx}[y] = \frac{dy}{dx}$.

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Solution

1. Solve for $\frac{dy}{dx}$ when $3x + 4y + 7 = 0$:

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}[3x + 4y + 7] = 0$$

Apply linear differentiation:

$$3 + 4\frac{dy}{dx} = 0$$

Rearrange to isolate $\frac{dy}{dx}$:

$$4\frac{dy}{dx} = -3 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{-3}{4}$$

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2. Solve for $\frac{dy}{dx}$ when $xy + y^2 = 2$:

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}[xy] + \frac{d}{dx}[y^2] = \frac{d}{dx}[2]$$

Apply the product rule to $xy$:

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

Group terms involving $\frac{dy}{dx}$:

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

Solve for $\frac{dy}{dx}$:

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

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3. Solve for $\frac{dy}{dx}$ when $x^2 - 4xy - 5y = 0$:

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}[x^2] - 4\frac{d}{dx}[xy] - 5\frac{d}{dx}[y] = 0$$

Apply the product rule to $xy$:

$$2x - 4\left(\frac{dy}{dx} \cdot x + y\right) - 5\frac{dy}{dx} = 0$$

Simplify and group $\frac{dy}{dx}$:

$$(4x + 5)\frac{dy}{dx} = 2x - 4y$$

Solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{2x - 4y}{4x + 5}$$

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4. Solve for $\frac{dy}{dx}$ when $4x^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$:

Differentiate both sides with respect to $x$:

$$8x + 2h\left(\frac{dy}{dx} \cdot x + y\right) + 2b y\frac{dy}{dx} + 2g + 2f\frac{dy}{dx} = 0$$

Group terms involving $\frac{dy}{dx}$:

$$2(hx + by + f)\frac{dy}{dx} = -2(4x + hy + g)$$

Solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{-2(4x + hy + g)}{2(hx + by + f)}$$

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5. Solve for $\frac{dy}{dx}$ when $x\sqrt{1+y} + y\sqrt{1+x} = 0$:

Differentiate both sides with respect to $x$, applying the product and chain rules:

$$\sqrt{1+y} + \frac{x}{2\sqrt{1+y}}\frac{dy}{dx} + \sqrt{1+x}\frac{dy}{dx} + \frac{y}{2\sqrt{1+x}} = 0$$

Group terms involving $\frac{dy}{dx}$:

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

Solve for $\frac{dy}{dx}$:

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

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6. Solve for $\frac{dy}{dx}$ when $y(x^2 - 1) = x\sqrt{x^2 + 4}$:

Differentiate both sides with respect to $x$:

$$\frac{dy}{dx}(x^2 - 1) + y(2x) = \sqrt{x^2 + 4} + \frac{x^2}{\sqrt{x^2 + 4}}$$

Group terms involving $\frac{dy}{dx}$:

$$(x^2 - 1)\frac{dy}{dx} = \sqrt{x^2 + 4} + \frac{x^2}{\sqrt{x^2 + 4}} - 2xy$$

Solve for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{\sqrt{x^2 + 4} + \frac{x^2}{\sqrt{x^2 + 4}} - 2xy}{x^2 - 1}$$

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Key Formulas or Methods Used

1. Chain Rule: For nested functions, differentiate the outer function first, then multiply by the derivative of the inner function.
2. Product Rule: $\frac{d}{dx}[uv] = u'v + uv'$.
3. Linear Differentiation: Constants and terms are differentiated directly.

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Summary of Steps

1. Differentiate the equation with respect to $x$, applying relevant rules.
2. Use the product and chain rules for terms involving $y$.
3. Group all $\frac{dy}{dx}$ terms on one side of the equation.
4. Solve for $\frac{dy}{dx}$ by isolating it.