Q.If $x = t^2$, $y = t^3$, then $\dfrac{d^2 y}{dx^2}$ is
(A) $\dfrac{3}{2}$
(B) $\dfrac{3}{4t}$
(C) $\dfrac{3}{2t}$
(D) $\dfrac{3}{4}$

The key insight here is to avoid jumping straight into parametric differentiation. When you see expressions like $\sec^2\theta$ and $\tan^2\theta$, recall the fundamental trigonometric identity: $\sec^2\theta - \tan^2\theta = 1$. This identity often lets you eliminate the parameter $\theta$ entirely, revealing a direct relationship between $x$ and $y$.

Let’s check:

$x = a \sec^2\theta$ and $y = a \tan^2\theta$.

Divide both equations by $a$: $\frac{x}{a} = \sec^2\theta$ and $\frac{y}{a} = \tan^2\theta$.

Subtract: $\frac{x}{a} - \frac{y}{a} = \sec^2\theta - \tan^2\theta = 1$.

So $x - y = a$, or $y = x - a$.

That’s a straight line with slope $1$ and intercept $-a$. Once you have $y$ as a linear function of $x$, the derivatives are trivial.

1. First derivative: Since $y = x - a$, differentiate with respect to $x$:

   $\frac{dy}{dx} = 1$.

2. Second derivative: Differentiate again:

   $\frac{d^2y}{dx^2} = 0$. …

We are given parametric equations:

We need $\frac{dy}{dx}$. Since both $x$ and $y$ are functions of $t$, we use the chain rule:

Concept & Intuition:

Parametric differentiation is straightforward: differentiate each coordinate with respect to the parameter $t$, then take the ratio. The only subtlety here is simplifying $dx/dt$, because it involves a logarithmic term with a trigonometric function inside. The trick is to rewrite $\log(\tan^2(t/2))$ as $2\log(\tan(t/2))$ and then differentiate carefully, using identities to simplify the result.

Step-by-step solution:

1. Differentiate $y$ with respect to $t$:

2. Differentiate $x$ with respect to $t$:

First, simplify the log term:

So

3. Differentiate term by term:

For the derivative of $\log(\tan(t/2))$, use the chain rule:

Since $\frac{1}{\tan} = \cot$, this becomes:

But $\cot \theta \sec^2 \theta = \frac{\cos\theta}{\sin\theta} \cdot \frac{1}{\cos^2\theta} = \frac{1}{\sin\theta \cos\theta} = \frac{2}{\sin 2\theta}$.

Here $\theta = t/2$, so $\sin 2\theta = \sin t$. Thus:

Therefore:

We are given $y = f(x)$, with $p = \frac{dy}{dx}$ and $q = \frac{d^2 y}{dx^2}$. We want $\frac{d^2 x}{dy^2}$.

Concept and intuition:

When we swap the roles of dependent and independent variables, derivatives transform in a non-trivial way. The first derivative is easy: $\frac{dx}{dy} = 1 / \frac{dy}{dx} = 1/p$. But the second derivative is not simply $1/q$; we must differentiate $1/p$ with respect to $y$, remembering that $p$ itself is a function of $x$, and $x$ is a function of $y$. This requires the chain rule.

Step-by-step reasoning:

1. First derivative swap Since $y = f(x)$ is invertible (locally, assuming $p \neq 0$), we have

2. Set up the second derivative We need

3. Apply the chain rule The variable $p$ is a function of $x$, and $x$ is a function of $y$. So

4. Differentiate $1/p$ with respect to $x$ Using the power rule and chain rule:

But $\frac{dp}{dx} = \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{d^2 y}{dx^2} = q$.

5. Substitute back So $$$$

Differentiate each coordinate with respect to $\theta$:

For $x=\cos\theta+\log\tan\frac{\theta}{2}$, using $\dfrac{d}{d\theta}\log\tan\frac{\theta}{2}=\dfrac{1}{\sin\theta}$,

Therefore

$\frac{dy}{dx}=\frac{\cos\theta}{\cos^2\theta/\sin\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta.$ …