Science:Math Exam Resources/Courses/MATH110/December 2011/Question 05 (a)

MATH110 December 2011

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Question 05 (a)
Differentiate the following function. $f(x)={\frac {\cos(x)}{\sin ^{2}(x)}}$

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
What differentiation rules will you need here?

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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. Since this is a quotient, we will be using the quotient rule. For this, we need to have the derivative of both the denominator and the numerator. The numerator is simply the cosine function and we know that: $\displaystyle (\cos(x))'=-\sin(x)$ For the denominator, we will need to use the chain rule and obtain: $\displaystyle \left(\sin ^{2}(x)\right)'=2\sin(x)\cos(x)$ So now we can use the quotient rule and obtain: ${\begin{aligned}f'(x)&={\frac {(\cos(x))'(\sin ^{2}(x))-(\cos(x))(\sin ^{2}(x))'}{(\sin ^{2}(x))^{2}}}\\&={\frac {(-\sin(x))(\sin ^{2}(x))-(\cos(x))(2\sin(x)\cos(x))}{\sin ^{4}(x)}}\\&={\frac {-\sin ^{3}(x)-2\sin(x)\cos ^{2}(x)}{\sin ^{4}(x)}}\\&={\frac {-\sin ^{2}(x)-2\cos ^{2}(x)}{\sin ^{3}(x)}}\\&=-{\frac {\sin ^{2}(x)+2\cos ^{2}(x)}{\sin ^{3}(x)}}\\&=-{\frac {1+\cos ^{2}(x)}{\sin ^{3}(x)}}\end{aligned}}$ Where the last equality was obtained using that $\sin ^{2}(x)+\cos ^{2}(x)=1$.

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