Science:Math Exam Resources/Courses/MATH102/December 2014/Question C 01

MATH102 December 2014

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Question C 01
Use implicit differentiation to calculate the derivative of $\theta (x)={\text{arccot}}(x)$. Recall that $\cot(x)={\frac {\cos(x)}{\sin(x)}}$.

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
Rewrite the expression as $x=\cot(\theta )={\frac {\cos(\theta )}{\sin(\theta )}}$ and use quotient rule ${\Big (}{\frac {g(x)}{h(x)}}{\Big )}'={\frac {g'(x)h(x)-g(x)h'(x)}{h^{2}(x)}}$ to calculate the implicit differentiation of the expression with respect to $x$.

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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 $x=\cot(\theta )$, we can rewrite it as $x={\frac {\cos(\theta )}{\sin(\theta )}},$ With implicit differentiation we take the derivative of above expression with respect to $x$ on both sides to get Graphs ${\begin{aligned}1&={\frac {-\sin ^{2}(\theta )-\cos ^{2}(\theta )}{\sin ^{2}(\theta )}}\cdot \theta ',\\1&=-{\frac {1}{\sin ^{2}(\theta )}}\cdot \theta '\end{aligned}}$ So we obtain $\theta '=-\sin ^{2}(\theta )=-{\frac {1}{1+x^{2}}},$ as shown in the following diagram Note: the triangle drawn on the right diagram can be solely constructed from the equation of $x=\cot(\theta )$

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