MATH100 December 2010
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q1 (g) • Q1 (h) • Q1 (i) • Q1 (j) • Q1 (k) • Q1 (l) • Q1 (m) • Q1 (n) • Q2 (a) • Q2 (b) • Q2 (c) • Q3 • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q4 (e) • Q5 • Q6 • Q7 • Q8 •
Question 01 (e)
Short-Answer Questions. Each question is worth 3 marks, but not all questions are of equal difficulty. Full marks will be given for correct answers placed in the box, but at most 1 mark will be given for incorrect answers. Unless otherwise stated, it is not necessary to simplify your answers in this question.
Find the derivative of
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.
Strap on your seal belts. We're going to need a lot of applications of the chain rule. If for differentiable f and g, then the chain rule states that
To help us out we should break the problem down into smaller subproblems. Let and let . Then . So the derivative by the chain rule is
To solve for the derivative of we again use the chain rule...
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Let and let . Then
So the derivative by the chain rule is
- (Equation 1)
Now, and so
- (Equation 2)
For , we again use the chain rule. Let and so we have that . Using the chain rule, we have that
- (Equation 3)
Notice that and thus
- (Equation 4)
Thus it suffices to find . Guess what we're going to use... the chain rule! Let and . Then, the chain rule states that
- (Equation 5)
Now, and so
- (Equation 6)
Here, the derivative of is actually doable! We have that
and hence by Equations 5 and 6, we have
- (Equation 7)
Combining Equations 1,2,3,4,6 and 7, we have
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