The image contains a mathematical problem, written in Hindi and mathematical notation. Let's break down the key elements.

The problem in part (क) states: $\text{यदि } e^{y}(x+1) = 1 \text{ तो दर्शाइये } \frac{d^2 y}{dx^2} = \left(\frac{dy}{dx}\right)^2$

This translates to:

• "If $e^{y}(x+1) = 1$, show that $\frac{d^2 y}{dx^2} = \left( \frac{dy}{dx} \right)^2$."

Solution:

Step 1: Start with the given equation

$e^y (x + 1) = 1$

Solve for $e^y$: $e^y = \frac{1}{x + 1}$

Step 2: Take the natural logarithm on both sides

Taking the natural logarithm $\ln$ on both sides gives: $y = \ln\left( \frac{1}{x + 1} \right)$ This simplifies to: $y = -\ln(x + 1)$

Step 3: First derivative (find $\frac{dy}{dx}$)

Now, differentiate $y = -\ln(x + 1)$ with respect to $x$: $\frac{dy}{dx} = -\frac{1}{x + 1}$

Step 4: Second derivative (find $\frac{d^2y}{dx^2}$)

Differentiate $\frac{dy}{dx} = -\frac{1}{x + 1}$ with respect to $x$: $\frac{d^2 y}{dx^2} = \frac{1}{(x + 1)^2}$

Step 5: Show the relation $\frac{d^2 y}{dx^2} = \left( \frac{dy}{dx} \right)^2$

From Step 3, we know that: $\frac{dy}{dx} = -\frac{1}{x + 1}$ Now, square both sides: $\left( \frac{dy}{dx} \right)^2 = \left( -\frac{1}{x + 1} \right)^2 = \frac{1}{(x + 1)^2}$

Thus, we have shown that: $\frac{d^2 y}{dx^2} = \left( \frac{dy}{dx} \right)^2$

Hence, the equation is verified.

Would you like further details or clarification on any step?

Here are 5 follow-up questions for deeper understanding:

1. What is the significance of differentiating a logarithmic function?
2. How can you apply this method to other similar problems involving exponentials?
3. What are some common techniques for solving second-order differential equations?
4. Can you explain how chain rule is used when differentiating implicitly?
5. How does logarithmic differentiation simplify problems involving exponents?

Tip: When differentiating logarithmic functions, always check if properties of logarithms can simplify your expressions before taking the derivative.