Q.If $y = \log\left(\dfrac{1 - x^2}{1 + x^2}\right)$, then $\dfrac{dy}{dx}$ is equal to
(A) $\dfrac{4x^3}{1 - x^4}$
(B) $\dfrac{-4x}{1 - x^4}$
(C) $\dfrac{1}{4 - x^4}$
(D) $\dfrac{-4x^3}{1 - x^4}$

Concept: Implicit Differentiation — but here we can directly differentiate using logarithm properties.

Step 1: Simplify using $\log\frac{a}{b} = \log a - \log b$:

Step 2: Differentiate term by term:

Step 3: Combine into a single fraction: …

Using $\log\left(\dfrac{A}{B}\right)=\log A-\log B$:

Differentiate term by term:

Combine the bracket: …

Method: Differentiating $y = \log\left(\dfrac{u(x)}{v(x)}\right)$ Using Log Laws First

Use this whenever the derivative you need is of a logarithm of a quotient (or product/power) — simplify the log expression algebraically before differentiating, rather than applying the quotient rule inside the log.

Steps

Step 1: Apply logarithm laws to break the single log into simpler pieces.

This single algebraic step is what makes the rest of the differentiation much cleaner.

Step 2: Differentiate each simplified log term separately using the standard rule.

$\frac{d}{dx}\log(u) = \frac{u'}{u}$ …

Common Mistakes

Mistake 1: Applying the quotient rule directly to $\frac{1-x^2}{1+x^2}$ inside the log, instead of splitting the log first using $\log(A/B) = \log A - \log B$.

Why it's wrong: this leads to a messier expression — $\frac{d}{dx}\log\left(\frac{1-x^2}{1+x^2}\right) = \frac{1+x^2}{1-x^2} \cdot \frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)$ — that is far more error-prone to simplify correctly than the split form.

Correct approach: always split $\log(A/B)$ into $\log A - \log B$ before differentiating; each piece then only needs the simple rule $\frac{d}{dx}\log(u)=\frac{u'}{u}$.

Mistake 2: A sign error when differentiating $\log(1-x^2)$, writing its derivative as $\frac{2x}{1-x^2}$ instead of $\frac{-2x}{1-x^2}$. …