Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>If $y = \sqrt{\dfrac{1-x}{1+x}}$, then $\dfrac{dy}{dx}$ equals:</p>
<p>$\dfrac{y}{x^2-1}$</p>
<p>$\dfrac{-y}{1-x^2}$</p>
<p>$\dfrac{y}{1+x^2}$</p>
<p>$\dfrac{y}{y^2-1}$</p>
Step-by-Step Solution
Key Concept: General
<b>Logarithmic Differentiation</b><br>Take $\ln$: $\ln y = \frac{1}{2}\ln(1-x) - \frac{1}{2}\ln(1+x)$<br>Differentiate: $\frac{1}{y}\frac{dy}{dx} = \frac{1}{2}\cdot\frac{-1}{1-x} - \frac{1}{2}\cdot\frac{1}{1+x} = -\frac{1}{1-x^2}$<br>$\therefore \frac{dy}{dx} = -\frac{y}{1-x^2}$<br><b>Key concept:</b> Logarithmic differentiation simplifies radical/power expressions.<br><b>Trap:</b> Sign error — both terms are negative, giving $-\frac{1}{1-x^2}$, not $+\frac{1}{1-x^2}$.
Correct Answer: B