Limits, Continuity & Differentiability
Methods of Differentiation
Grade 12
Question:
<p>If $f(x) = \begin{cases} x\tan^{-1}(1/x), & x \in (-\infty,-1]\cup[1,\infty) \\ 0, & x = 0 \end{cases}$, then $f'(0)$ is:</p>
<p>(B) equal to 0</p>
<p>(C) equal to 1</p>
<p>(D) non-existent</p>
<p>(A) -1</p>
Step-by-Step Solution
Key Concept: General
<b>Checking Differentiability at x = 0</b><br>$f'(0) = \lim_{h\to0}\frac{f(h)-f(0)}{h} = \lim_{h\to0}\frac{h\tan^{-1}(1/h)}{h} = \lim_{h\to0}\tan^{-1}(1/h)$<br>As $h\to0^+$: $\tan^{-1}(1/h) \to \pi/2$<br>As $h\to0^-$: $\tan^{-1}(1/h) \to -\pi/2$<br>LHD $\ne$ RHD, so $f'(0)$ does not exist.<br><b>Key concept:</b> Derivative exists iff LHD = RHD.<br><b>Trap:</b> $f(0)=0$ does NOT mean $f'(0)=0$; the limit of the difference quotient must be computed.
Correct Answer: A