Q.If $x = e^{\cos 2t}$ and $y = e^{\sin 2t}$, prove that $\dfrac{dy}{dx} = -\dfrac{y \log x}{x \log y}$.

Concept: Implicit Differentiation (parametric form, then using the given relation).

Step 1 – Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

$x = e^{\cos 2t} \implies \log x = \cos 2t$

Differentiating: $\frac{1}{x}\frac{dx}{dt} = -2\sin 2t \implies \frac{dx}{dt} = -2x\sin 2t$

$y = e^{\sin 2t} \implies \log y = \sin 2t$

Differentiating: $\frac{1}{y}\frac{dy}{dt} = 2\cos 2t \implies \frac{dy}{dt} = 2y\cos 2t$

Step 2 – Compute $\frac{dy}{dx}$

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2y\cos 2t}{-2x\sin 2t} = -\frac{y}{x}\cot 2t$

Step 3 – Express $\cot 2t$ in terms of $x$ and $y$ …

We are given two parametric equations:

$x = e^{\cos 2t}$ and $y = e^{\sin 2t}$.

We need to prove that $\frac{dy}{dx} = -\frac{y \log x}{x \log y}$.

The direct approach — differentiating $x$ and $y$ with respect to $t$ and then using $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ — is tempting. But notice the result involves $\log x$ and $\log y$, which are not directly in the parametric forms. This suggests we should first eliminate the parameter $t$ by taking natural logarithms, then use implicit differentiation on the resulting relation between $x$ and $y$.

Why implicit differentiation?

Implicit differentiation lets us find $\frac{dy}{dx}$ without solving for $y$ explicitly in terms of $x$. Here, after taking logs, we get a simple relation: $\cos 2t = \log x$ and $\sin 2t = \log y$. Squaring and adding gives $\log^2 x + \log^2 y = 1$, which is an implicit equation linking $x$ and $y$. Differentiating this directly yields the required derivative.

Let’s work through it step by step.

1. Take natural logarithms of both parametric equations.

   From $x = e^{\cos 2t}$, we have $\log x = \cos 2t$.

   From $y = e^{\sin 2t}$, we have $\log y = \sin 2t$.

   (Here $\log$ denotes the natural logarithm, base $e$.)

2. Eliminate $t$ by squaring and adding.

   $\log^2 x + \log^2 y = \cos^2 2t + \sin^2 2t = 1$.

   So the relation between $x$ and $y$ is:

3. Differentiate both sides implicitly with respect to $x$.

   Remember that $y$ is a function of $x$. Differentiate term by term:

   • Derivative of $(\log x)^2$: $2(\log x) \cdot \frac{1}{x}$.
   • Derivative of $(\log y)^2$: $2(\log y) \cdot \frac{1}{y} \cdot \frac{dy}{dx}$ (by the chain rule).
   • Derivative of the constant $1$ is $0$.

   So we get:

4. Solve for $\frac{dy}{dx}$. Divide through by $2$:

Rearranging:

$\frac{\log y}{y} \cdot \frac{dy}{dx} = -\frac{\log x}{x}.$ …

Method: Eliminate the Parameter via Logarithms, Then Differentiate Implicitly

Use this when $x$ and $y$ are both given as exponentials of a trig function of the same parameter (e.g. $x = e^{\cos 2t}$, $y = e^{\sin 2t}$) and the result you must reach is expressed in terms of $\log x$ and $\log y$ rather than the parameter itself — this is a strong hint that eliminating $t$ before differentiating is faster than differentiating through $t$ first.

Steps

Step 1: Take logarithms of both parametric equations

Since $x = e^{f(t)}$, taking $\log$ of both sides gives simply $\log x = f(t)$ (the exponential and logarithm undo each other). Do the same for $y$.

Step 2: Use a Pythagorean-style identity to eliminate $t$

If the two resulting expressions are $\cos(\cdot)$ and $\sin(\cdot)$ of the same angle, squaring both and adding uses $\sin^2+\cos^2=1$ to produce a single equation relating $\log x$ and $\log y$ directly — no $t$ left at all.

Step 3: Differentiate the resulting equation implicitly with respect to $x$ …

Common Mistakes

Mistake 1: Dropping the chain-rule factor of $2$ when differentiating $e^{\cos2t}$

Why it's wrong: writing $\dfrac{dx}{dt}=e^{\cos2t}\cdot(-\sin2t)$ forgets to also multiply by the derivative of the inner $2t$, i.e. the extra factor of $2$. Correct approach: $\dfrac{d}{dt}\cos2t=-2\sin2t$, so $\dfrac{dx}{dt}=-2x\sin2t$.

Mistake 2: Sign slip on $\dfrac{d}{dt}(\cos2t)$

Why it's wrong: dropping the negative sign here flips the sign of the whole final ratio, turning a correct $-\tfrac{y\log x}{x\log y}$ into its positive counterpart. Correct approach: keep track that $\cos$ differentiates to $-\sin$, always.

Mistake 3: Swapping $\log x$ and $\log y$ when substituting back …