Q.Find $\dfrac{dy}{dx}$ of the function expressed in parametric form: $x = e^{\theta}\left(\theta + \dfrac{1}{\theta}\right),\ y = e^{-\theta}\left(\theta - \dfrac{1}{\theta}\right)$.

Concept: Parametric differentiation — $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$.

Differentiate $x=e^{\theta}\left(\theta+\frac1\theta\right)$ (product rule):

Differentiate $y=e^{-\theta}\left(\theta-\frac1\theta\right)$:

$\frac{dy}{d\theta}=-e^{-\theta}\left(\theta-\frac1\theta\right)+e^{-\theta}\left(1+\frac{1}{\theta^2}\right)=e^{-\theta}\left(-\theta+\frac1\theta+1+\frac{1}{\theta^2}\right).$ …

The idea

Both $x$ and $y$ are given through the parameter $\theta$, so we cannot write $y$ directly in terms of $x$. Instead use

Step-by-step

1. $\dfrac{dx}{d\theta}$. For $x=e^{\theta}\left(\theta+\frac1\theta\right)$, product rule:

2. $\dfrac{dy}{d\theta}$. For $y=e^{-\theta}\left(\theta-\frac1\theta\right)$, product rule (note $\frac{d}{d\theta}e^{-\theta}=-e^{-\theta}$):

3. Divide. The exponentials give $e^{-\theta}/e^{\theta}=e^{-2\theta}$:

$\frac{dy}{dx}=e^{-2\theta}\cdot\frac{-\theta+\frac1\theta+1+\frac{1}{\theta^2}}{\theta+\frac1\theta+1-\frac{1}{\theta^2}}.$ …

Method: Parametric Differentiation When the Functions Are Products

Use this when $x(t)$ and/or $y(t)$ is written as a product of two simpler functions of $t$ (e.g. an exponential times a polynomial or a rational expression) — you need the Product Rule inside the parametric-differentiation process.

Steps

Step 1: Identify the two factors in each product

For an expression like $e^{t}\cdot h(t)$, label $u = e^{t}$ and $v = h(t)$ so the Product Rule $\dfrac{d}{dt}(uv) = u'v + uv'$ can be applied cleanly.

Step 2: Differentiate $x$ using the Product Rule

Keep the two resulting terms separate at first rather than trying to combine them immediately — combining too early causes sign errors.

Step 3: Differentiate $y$ the same way

Repeat for $y(t)$, being careful with the derivative of $e^{-t}$ (it carries a negative sign, unlike $e^{t}$).

Step 4: Form the ratio $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ …

Common Mistakes

Mistake 1: Forgetting the product rule on $e^{\theta}\left(\theta+\frac1\theta\right)$

Why it's wrong: differentiating only the second factor and writing $\dfrac{dx}{d\theta}=e^{\theta}\left(1-\tfrac{1}{\theta^2}\right)$ drops the term coming from differentiating $e^\theta$ itself. Correct approach: apply the product rule fully — $\dfrac{dx}{d\theta}=e^{\theta}\left(\theta+\tfrac1\theta\right)+e^{\theta}\left(1-\tfrac{1}{\theta^2}\right)$.

Mistake 2: Dropping the negative sign on $\dfrac{d}{d\theta}e^{-\theta}$

Why it's wrong: treating $e^{-\theta}$'s derivative as $+e^{-\theta}$ instead of $-e^{-\theta}$ inverts the sign of the first term in $dy/d\theta$, which changes the whole final ratio. Correct approach: always carry the chain-rule factor of $-1$ that comes from differentiating the exponent $-\theta$.

Mistake 3: Combining the $\theta^2$-fractions incorrectly before dividing …