Science:Math Exam Resources/Courses/MATH100/December 2016/Question 09 (b) (i)

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MATH100 December 2016

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Question 09 (b) (i)
Let $f(x)={\frac {\sqrt {x^{2}-8}}{x-4}}$ . (i) Compute and simplify $f'(x)$ .

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Use the quotient rule, or logarithmic differentiation.

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Solution
We find $f'(x)$ on $(\mathbb {R} \setminus [-2{\sqrt {2}},2{\sqrt {2}}])\setminus \{4\}.$ Note that on this set, $f(x)$ is well-defined and doesn't vanishes. (i.e., $f(x)\neq 0.$ .) To do so, we use logarithmic differentiation. For this purpose, define $g$ by $g(x)=\ln \|f(x)\|={\frac {1}{2}}\ln \|x^{2}-8\|-\ln \|x-4\|$ . Then, its derivative can be easily calculated, and is given by $g'(x)={\frac {1}{2}}\cdot {\frac {1}{x^{2}-8}}\cdot 2x-{\frac {1}{x-4}}={\frac {x}{x^{2}-8}}-{\frac {1}{x-4}}$ . On the other hand, by the chain rule, $g'$ relates with $f'$ in the following way: $f'(x)=f(x)g'(x)={\frac {\sqrt {x^{2}-8}}{x-4}}\left({\frac {x}{x^{2}-8}}-{\frac {1}{x-4}}\right).$ Hence for any $x\in \{x\in (\mathbb {R} \setminus [-2{\sqrt {2}},2{\sqrt {2}}])\setminus \{4\},$ $f'(x)=\color {blue}{\frac {\sqrt {x^{2}-8}}{x-4}}\left({\frac {x}{x^{2}-8}}-{\frac {1}{x-4}}\right)$ .