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

Concept: Implicit Differentiation — differentiate both sides with respect to $x$, treating $y$ as a function of $x$.

Step 1: Differentiate $e^y (x+1) = 1$ implicitly:

Factor $e^y$:

Since $e^y \neq 0$, we get:

Step 2: Differentiate again with respect to $x$:

$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( -\frac{1}{x+1} \right) = \frac{1}{(x+1)^2}$ …

We start with the equation $e^y (x+1) = 1$. This is an implicit relation between $x$ and $y$ — we cannot easily solve for $y$ in terms of $x$ (though here we could, but implicit differentiation is cleaner). The goal is to prove that the second derivative $\frac{d^2 y}{dx^2}$ equals $\left(\frac{dy}{dx}\right)^2$.

The intuition: implicit differentiation lets us differentiate both sides with respect to $x$, treating $y$ as a function of $x$. Every time we differentiate $y$, we multiply by $\frac{dy}{dx}$ (chain rule). Then we differentiate again, and the algebra should collapse nicely.

Let’s work through it.

1. Differentiate once. Given: $e^y (x+1) = 1$. Differentiate both sides with respect to $x$:

Use the product rule:

Now $\frac{d}{dx}(e^y) = e^y \frac{dy}{dx}$ (chain rule), and $\frac{d}{dx}(x+1) = 1$. So:

Factor $e^y$:

Since $e^y > 0$ for all real $y$, we can divide by $e^y$:

So the first derivative is:

2. Differentiate again. We need $\frac{d^2 y}{dx^2}$. Differentiate the equation $(x+1) \frac{dy}{dx} + 1 = 0$ with respect to $x$:

Use the product rule on the first term:

That is:

So:

Rearranging:

$\frac{d^2 y}{dx^2} = -\frac{1}{x+1} \cdot \frac{dy}{dx}.$ …

Method: Implicit Differentiation Twice, Substituting Back

For an equation relating $x$ and $y$ (with no way to isolate $y$ cleanly, or where it's cleaner not to), differentiate implicitly to get $y_1$ in terms of $x$ (and possibly $y$), then differentiate that equation again — and substitute the first-derivative relation back in to simplify.

Steps

Step 1: Differentiate the given relation implicitly to find $y_1$

Use the product rule and chain rule as needed, remembering every $y$-term picks up a factor of $y_1$.

Step 2: Differentiate the Step-1 equation again (not just the isolated $y_1$ formula) to bring in $y_2$

Step 3: Substitute the known value/expression for $y_1$ back into the $y_2$ equation to reach the final simplified relation …

Common Mistakes

Mistake 1: Forgetting the product rule when differentiating $(x+1)y_1$ a second time.

Why it's wrong: $(x+1)y_1$ is a product of two functions of $x$ (both $(x+1)$ and $y_1$ vary), so its derivative needs the product rule — treating $(x+1)$ as if it were a fixed constant on the second pass loses the $y_1$ term entirely. Correct approach: write $\frac{d}{dx}[(x+1)y_1]=y_1+(x+1)y_2$ explicitly.

Mistake 2: Not substituting the first-derivative expression back into the second-derivative equation. …