Q.Find $\dfrac{dy}{dx}$ of the function expressed in parametric form: $x = \dfrac{1 + \log t}{t^2},\ y = \dfrac{3 + 2\log t}{t}$.

For a parametric curve, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.

Differentiate $x=\dfrac{1+\log t}{t^2}$ by the quotient rule:

Differentiate $y=\dfrac{3+2\log t}{t}$:

$\frac{dy}{dt}=\frac{\frac{2}{t}\cdot t-(3+2\log t)}{t^2}=\frac{-1-2\log t}{t^2}.$ …

A curve in parametric form gives $x$ and $y$ separately as functions of a parameter $t$. To get the slope we use

because both derivatives are taken with respect to the same $t$, so the $dt$'s cancel.

1. Differentiate $x$

With $x=\dfrac{1+\log t}{t^2}$, the quotient rule gives

2. Differentiate $y$

With $y=\dfrac{3+2\log t}{t}$, …

Method: Parametric Differentiation with Quotient-Rule and Logarithmic Terms

Use this when $x(t)$ and/or $y(t)$ is a fraction whose numerator involves $\log t$ — you need the Quotient Rule for each derivative, and the two results typically share a common factor that cancels in the final ratio.

Steps

Step 1: Apply the Quotient Rule to $x(t)$

For $x = \dfrac{p(t)}{q(t)}$,

Remember $\dfrac{d}{dt}\log t = \dfrac{1}{t}$ when differentiating the numerator.

Step 2: Apply the Quotient Rule to $y(t)$ the same way

Simplify the numerator of each result as far as possible before moving on — expand any products and collect like terms.

Step 3: Form the ratio $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ …

Common Mistakes

Mistake 1: Misapplying the quotient rule's numerator order

Why it's wrong: for $x=\dfrac{1+\log t}{t^2}$, writing the numerator as $t^2\cdot(1+\log t)'-(1+\log t)\cdot(t^2)'$ reversed, or forgetting the $(t^2)^2=t^4$ denominator, gives a wrong sign or wrong power of $t$. Correct approach: apply $\left(\tfrac{u}{v}\right)'=\tfrac{u'v-uv'}{v^2}$ carefully with $u=1+\log t,\ v=t^2$.

Mistake 2: Using the wrong derivative for $\log t$

Why it's wrong: treating $\dfrac{d}{dt}(\log t)$ as $1$ instead of $\dfrac1t$ throws off both $dx/dt$ and $dy/dt$ from the start. Correct approach: always keep $\dfrac{d}{dt}(\log t)=\dfrac1t$ explicit before combining terms.

Mistake 3: Missing the common-factor cancellation …