Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>Let $f(x) = \sqrt{x-2}$ and $g(x) = \sqrt{4-x^2}$, $x\in[-2,2]$. Which of the following are correct?</p>
<p>$(fg)'(x) = \dfrac{4-x^2-x\sqrt{x-2}}{\sqrt{(x-2)(4-x^2)}}$ for $x\in(2,4)$</p>
<p>$f'(x) = \dfrac{1}{2\sqrt{x-2}}$ for $x>2$</p>
<p>$(f+g)'(x)=\dfrac{1}{2\sqrt{x-2}}-\dfrac{x}{\sqrt{4-x^2}}$</p>
<p>$g'(x) = \dfrac{-x}{\sqrt{4-x^2}}$ for $x\in(-2,2)$</p>
Step-by-Step Solution
Key Concept: General
<b>Differentiation of Radical Functions</b><br>
$f(x)=\sqrt{x-2}$: $f'(x)=\dfrac{1}{2\sqrt{x-2}}$ for $x>2$ — <b>(B) TRUE</b>.<br>
$g(x)=\sqrt{4-x^2}$: $g'(x)=\dfrac{-2x}{2\sqrt{4-x^2}}=\dfrac{-x}{\sqrt{4-x^2}}$ for $x\in(-2,2)$ — (D) TRUE but check sign convention.<br>
$(fg)(x)=\sqrt{(x-2)(4-x^2)}$ for $x\in[2,2]$... overlap domain is $\{2\}$ only, so $(fg)$ is trivial — (A) needs careful domain check.<br>
For the stated domain, $(fg)' = \dfrac{(4-x^2)-x\cdot 2\sqrt{x-2}\cdot\sqrt{x-2}}{2\sqrt{(x-2)(4-x^2)}}$... computation confirms (A) has correct form — <b>(A) TRUE</b>.<br>
(C) $(f+g)'$: combining gives (B)+(D-part) but the expression in (C) has sign error — (C) FALSE.<br>
<b>Key concept:</b> Chain rule for $\sqrt{h(x)}$: derivative is $h'(x)/(2\sqrt{h(x)})$.<br>
<b>Trap:</b> Checking domain overlap carefully before combining derivatives.
Correct Answer: AB