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 •

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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 $\displaystyle y=\sin ^{5}(\cos(3x^{2}))$

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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
Strap on your seal belts. We're going to need a lot of applications of the chain rule. If $h(x)=f(g(x))$ for differentiable f and g, then the chain rule states that $\displaystyle h'(x)=f'(g(x))g'(x)$

Hint 2
To help us out we should break the problem down into smaller subproblems. Let $f_{1}(x)=x^{5}$ and let $f_{2}(x)=\sin(\cos(3x^{2}))$ . Then $y=f_{1}(f_{2}(x))=\sin ^{5}(\cos(3x^{2}))$ . So the derivative by the chain rule is $\displaystyle y'=f'_{1}(f_{2}(x))f_{2}'(x)$ To solve for the derivative of $f_{2}(x)$ we again use the chain rule...

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Let $\displaystyle f_{1}(x)=x^{5}$ and let $\displaystyle f_{2}(x)=\sin(\cos(3x^{2}))$ . Then $\displaystyle y=f_{1}(f_{2}(x))=\sin ^{5}(\cos(3x^{2}))$ . So the derivative by the chain rule is $\displaystyle y'=f'_{1}(f_{2}(x))f_{2}'(x)$ (Equation 1) Now, $\displaystyle f'_{1}(x)=5x^{4}$ and so $\displaystyle f'_{1}(f_{2}(x))=5(\sin(\cos(3x^{2})))^{4}=5\sin ^{4}(\cos(3x^{2}))$ (Equation 2) For $\displaystyle f'_{2}(x)$ , we again use the chain rule. Let $\displaystyle f_{3}(x)=\sin(x)$ and $\displaystyle f_{4}(x)=\cos(3x^{2})$ so we have that $\displaystyle f_{2}(x)=f_{3}(f_{4}(x))$ . Using the chain rule, we have that $\displaystyle f'_{2}(x)=f'_{3}(f_{4}(x))f_{4}'(x)$ (Equation 3) Notice that $\displaystyle f_{3}'(x)=\cos(x)$ and thus $\displaystyle f'_{3}(f_{4}(x))=\cos(\cos(3x^{2}))$ (Equation 4) Thus it suffices to find $\displaystyle f'_{4}(x)$ . Guess what we're going to use... the chain rule! Let $\displaystyle f_{5}(x)=\cos(x)$ and $\displaystyle f_{6}(x)=3x^{2}$ . Then, the chain rule states that $\displaystyle f'_{4}(x)=f'_{5}(f_{6}(x))f'_{6}(x)$ (Equation 5) Now, $\displaystyle f'_{5}(x)=-\sin(x)$ and so $\displaystyle f'_{5}(f_{6}(x))=-\sin(3x^{2})$ (Equation 6) Here, the derivative of $\displaystyle f_{6}(x)$ is actually doable! We have that $\displaystyle f'_{6}(x)={\frac {d}{dx}}(3x^{2})=6x$ and hence by Equations 5 and 6, we have $\displaystyle f'_{4}(x)=f'_{5}(f_{6}(x))f'_{6}(x)=(-\sin(3x^{2}))(6x)$ (Equation 7) Combining Equations 1,2,3,4,6 and 7, we have ${\begin{aligned}y'&=f'_{1}(f_{2}(x))f_{2}'(x)&&{\text{(via Equation 1)}}\\&=5\sin ^{4}(\cos(3x^{2}))(f'_{3}(f_{4}(x))f_{4}'(x))&&{\text{(via Equation 2 and 3)}}\\&=5\sin ^{4}(\cos(3x^{2}))(\cos(\cos(3x^{2}))(-\sin(3x^{2}))(6x)))&&{\text{(via Equation 4 and 7)}}\\&=-30x\sin ^{4}(\cos(3x^{2}))\cos(\cos(3x^{2}))\sin(3x^{2}))\end{aligned}}$ Phew!