3.1 Q-2

Question Statement

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

1. $xy + x = 4$
2. $x^2 + 2y^2 = 16$
3. $x^4 + y^2 = xy^2$
4. $xy - \ln x = c$

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Background and Explanation

To solve these problems, we use the concept of implicit differentiation. This method involves differentiating both sides of an equation with respect to one variable (usually $x$), treating other variables (like $y$) as functions of $x$. Key rules include:

1. Product Rule: $d(uv) = u , dv + v , du$
2. Chain Rule: Differentiating $y$ with respect to $x$ gives $\frac{dy}{dx}$.

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Solution

1. Equation: $xy + x = 4$

Step-by-step:

1. Differentiate both sides:

$$d(xy + x) = d(4)$$

Using the product rule for $xy$:

$$d(xy) + dx = 0 \quad \Rightarrow \quad x , dy + y , dx + dx = 0$$

2. Rearrange to isolate $dy$:

$$x , dy = -y , dx - dx \quad \Rightarrow \quad dy = -\frac{(y+1)}{x} dx$$

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

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

4. Find its reciprocal:

$$\frac{dx}{dy} = -\frac{x}{y+1} \quad \text{(valid for $y \neq -1$)}.$$

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2. Equation: $x^2 + 2y^2 = 16$

Step-by-step:

1. Differentiate both sides:

$$d(x^2) + d(2y^2) = d(16) \quad \Rightarrow \quad 2x , dx + 4y , dy = 0$$

2. Rearrange to isolate $dy$:

$$4y , dy = -2x , dx \quad \Rightarrow \quad dy = -\frac{x}{2y} dx$$

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

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

4. Find its reciprocal:

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

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3. Equation: $x^4 + y^2 = xy^2$

Step-by-step:

1. Differentiate both sides:

$$d(x^4) + d(y^2) = d(xy^2) \quad \Rightarrow \quad 4x^3 , dx + 2y , dy = x , d(y^2) + y^2 , dx$$

Using the product rule on $xy^2$:

$$4x^3 , dx + 2y , dy = 2xy , dy + y^2 , dx$$

2. Rearrange to isolate $dy$:

$$2y , dy - 2xy , dy = y^2 , dx - 4x^3 , dx$$ $$dy \left( 2y - 2xy \right) = dx \left( y^2 - 4x^3 \right)$$ $$dy = \frac{y^2 - 4x^3}{2y(1-x)} dx$$

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

$$\frac{dy}{dx} = \frac{y^2 - 4x^3}{2y(1-x)}$$

4. Find its reciprocal:

$$\frac{dx}{dy} = \frac{2y(x-1)}{4x^3 - y^2}.$$

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4. Equation: $xy - \ln x = c$

Step-by-step:

1. Differentiate both sides:

$$d(xy) - d(\ln x) = d(c) \quad \Rightarrow \quad x , dy + y , dx - \frac{1}{x} , dx = 0$$

2. Rearrange to isolate $dy$:

$$x , dy = -y , dx + \frac{1}{x} , dx \quad \Rightarrow \quad dy = \frac{\frac{1}{x} - y}{x} dx$$

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

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

4. Find its reciprocal:

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

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Key Formulas or Methods Used

1. Product Rule: $d(uv) = u , dv + v , du$
2. Chain Rule: $\frac{d}{dx}[y] = \frac{dy}{dx}$
3. Implicit Differentiation: Differentiate with respect to one variable while treating others as functions of it.

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Summary of Steps

1. Differentiate the given equation using implicit differentiation.
2. Apply product and chain rules wherever necessary.
3. Rearrange the equation to isolate $dy/dx$ (or $dx/dy$).
4. Simplify the result to find the required derivative.

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