Science:Math Exam Resources/Courses/MATH100/December 2012/Question 09

MATH100 December 2012

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Question 09
Full-Solution Problem. In this question, justify your answer and show all your work. Simplification of numerical answers is required. Using the definition of the derivative, compute $\displaystyle f'(x)$ if $f(x)={\sqrt {1-2x^{2}}}$. Also, specify the domain of $\displaystyle f'(x)$. No marks will be given for use of differentiation rules, but you may use them to check your answer.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
A reminder that there are two equivalent definitions of the derivative. You can use either $f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$ or $f'(a)=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}$

Hint 2
You at some point will want to consider rationalizing the numerator.

Hint 3
To find the domain consider which fractions do not make sense and where one cannot evaluate square roots.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We proceed directly ${\begin{aligned}f'(x)&=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}\\&=\lim _{h\to 0}{\frac {{\sqrt {1-2(x+h)^{2}}}-{\sqrt {1-2x^{2}}}}{h}}\\&=\lim _{h\to 0}\left({\frac {{\sqrt {1-2(x+h)^{2}}}-{\sqrt {1-2x^{2}}}}{h}}\right)\cdot \left({\frac {{\sqrt {1-2(x+h)^{2}}}+{\sqrt {1-2x^{2}}}}{{\sqrt {1-2(x+h)^{2}}}+{\sqrt {1-2x^{2}}}}}\right)\\&=\lim _{h\to 0}{\frac {1-2(x+h)^{2}-(1-2x^{2})}{h({\sqrt {1-2(x+h)^{2}}}+{\sqrt {1-2x^{2}}})}}\\&=\lim _{h\to 0}{\frac {1-2(x^{2}+2xh+h^{2})-1+2x^{2}}{h({\sqrt {1-2(x+h)^{2}}}+{\sqrt {1-2x^{2}}})}}\\&=\lim _{h\to 0}{\frac {1-2x^{2}-4xh-2h^{2}-1+2x^{2}}{h({\sqrt {1-2(x+h)^{2}}}+{\sqrt {1-2x^{2}}})}}\\&=\lim _{h\to 0}{\frac {-4xh-2h^{2}}{h({\sqrt {1-2(x+h)^{2}}}+{\sqrt {1-2x^{2}}})}}\\&=\lim _{h\to 0}{\frac {-4x-2h}{{\sqrt {1-2(x+h)^{2}}}+{\sqrt {1-2x^{2}}}}}\\&={\frac {-4x-2(0)}{{\sqrt {1-2(x+(0))^{2}}}+{\sqrt {1-2x^{2}}}}}\\&={\frac {-4x}{{\sqrt {1-2x^{2}}}+{\sqrt {1-2x^{2}}}}}\\&={\frac {-4x}{2{\sqrt {1-2x^{2}}}}}\\&={\frac {-2x}{\sqrt {1-2x^{2}}}}\end{aligned}}$ Lastly, the domain of this function is anywhere that $\displaystyle 1-2x^{2}>0$ as square roots cannot be evaluated for negative numbers and if the quantity were zero, then the derivative would have a zero denominator and so we want that the term inside the radical is strictly positive. This occurs when $\displaystyle -1/{\sqrt {2}}<x<1/{\sqrt {2}}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We could also have used the other formula ${\begin{aligned}f'(a)&=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}\\&=\lim _{x\to a}{\frac {{\sqrt {1-2x^{2}}}-{\sqrt {1-2a^{2}}}}{x-a}}\\&=\lim _{x\to a}\left({\frac {{\sqrt {1-2x^{2}}}-{\sqrt {1-2a^{2}}}}{x-a}}\right)\cdot \left({\frac {{\sqrt {1-2x^{2}}}+{\sqrt {1-2a^{2}}}}{{\sqrt {1-2x^{2}}}+{\sqrt {1-2a^{2}}}}}\right)\\&=\lim _{x\to a}{\frac {1-2x^{2}-(1-2a^{2})}{(x-a)({\sqrt {1-2x^{2}}}+{\sqrt {1-2a^{2}}})}}\\&=\lim _{x\to a}{\frac {1-2x^{2}-1+2a^{2}}{(x-a)({\sqrt {1-2x^{2}}}+{\sqrt {1-2a^{2}}})}}\\&=\lim _{x\to a}{\frac {-2x^{2}+2a^{2}}{(x-a)({\sqrt {1-2x^{2}}}+{\sqrt {1-2a^{2}}})}}\\&=\lim _{x\to a}{\frac {-2(x^{2}-a^{2})}{(x-a)({\sqrt {1-2x^{2}}}+{\sqrt {1-2a^{2}}})}}\\&=\lim _{x\to a}{\frac {-2(x-a)(x+a)}{(x-a)({\sqrt {1-2x^{2}}}+{\sqrt {1-2a^{2}}})}}\\&=\lim _{x\to a}{\frac {-2(x+a)}{{\sqrt {1-2x^{2}}}+{\sqrt {1-2a^{2}}}}}\\&={\frac {-2(a+a)}{{\sqrt {1-2a^{2}}}+{\sqrt {1-2a^{2}}}}}\\&={\frac {-4a}{2{\sqrt {1-2a^{2}}}}}\\&={\frac {-2a}{\sqrt {1-2a^{2}}}}\end{aligned}}$ The domain is computed as in solution 1.

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