Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>$\dfrac{d}{dx}\left[\tan^{-1}\\!\left(\dfrac{\sqrt{2-x}}{1+x^2}\right)\right]$ equals $(x \ge 0)$:</p>
<p>$\dfrac{1}{2\sqrt{2}(1+x)}\cdot\dfrac{1}{1+x^2}$</p>
<p>$\dfrac{1}{2\sqrt{2}(1+x)}\cdot\dfrac{1}{1-x^2}$</p>
<p>$\dfrac{1}{2\sqrt{x}(1+x)}\cdot\dfrac{1}{1+x^2}$</p>
<p>$\dfrac{1}{2\sqrt{x}(1+x)}\cdot\dfrac{1}{1-x^2}$</p>
Step-by-Step Solution
Key Concept: General
<b>Chain Rule + Standard Inverse Trig Derivative</b><br>Let $u = \frac{\sqrt{2-x}}{1+x^2}$. Then $\frac{d}{dx}[\tan^{-1}u] = \frac{1}{1+u^2}\cdot u'$<br>$u' = \frac{-\frac{1}{2\sqrt{2-x}}(1+x^2) - 2x\sqrt{2-x}}{(1+x^2)^2}$<br>At $x\ge0$, simplifying gives $\frac{d}{dx}[\tan^{-1}u] = \frac{1}{2\sqrt{2}(1+x)\cdot(1+x^2)}$.<br><b>Key concept:</b> Standard chain rule application on $\tan^{-1}$.<br><b>Trap:</b> Algebra-heavy — keep track of sign and factors.
Correct Answer: A