Science:Math Exam Resources/Courses/MATH110/April 2018/Question 01 (e)

MATH110 April 2018

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Question 01 (e)
Find ${\textstyle y'}$ if ${\textstyle y=\sin ^{2}(2\pi x)}$ .

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
Write ${\textstyle \sin ^{2}(2\pi x)=\sin(2\pi x)\cdot \sin(2\pi x)}$ and use the product rule.

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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. Let ${\textstyle g(x)=\sin(2\pi x)}$ . By the product rule, we know that $y'=g'(x)\cdot g(x)+g(x)\cdot g'(x)=2g'(x)\cdot g(x),$ so we just have to find ${\textstyle g'(x)}$ . Note that $g$ can be written as a composite of two functions: $g(x)=h(w(x))$ where $h(y)=\sin(y)$ and $w(x)=2\pi x$ . By the chain rule with $h'(y)=\cos(y)$ and $w'(x)=2\pi$ , we have $g'(x)=h'(w(x))\cdot w'(x)=2\pi \cos(2\pi x).$ It follows that ${\textstyle y'=2\cdot 2\pi \cos(2\pi x)\cdot \sin(2\pi x)=4\pi \sin(2\pi x)\cos(2\pi x).}$ We can use the double-angle formula ${\textstyle \sin(2\theta )=2\sin \theta \cos \theta }$ to simplify this further. We get ${\textstyle y'=2\pi \sin(4\pi x).}$ Answer: The answer is ${\textstyle {\color {blue}y'=2\pi \sin(4\pi x)}}$ .

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 can write $y$ as a composite of two functions, $y=f(g(x)),\quad f(u)=u^{2},\quad g(x)=\sin(2\pi x).$ To find $y'$ , we apply the chain rule with $f'(u)=2u$ . Then, we get $y'=f'(g(x))g'(x)=2g(x)\cdot g'(x).$ By Solution 1, we have $g'(x)=2\pi \cos(2\pi x)$ , so that the answer is ${\textstyle {\color {blue}y'=2\pi \sin(4\pi x)}}$ .