Science:Math Exam Resources/Courses/MATH100/December 2010/Question 01 (e)

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

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
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!

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