Science:Math Exam Resources/Courses/MATH100/December 2012/Question 02 (c)

MATH100 December 2012

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Question 02 (c)
Short-Answer Questions. Questions 1-4 are short-answer questions. Put your answers in the boxes provided. Simplify your answers as much as possible, and show your work. Each question is worth 3 marks, but not all questions are of equal difficulty. If $\displaystyle y=\sin ^{-1}(\ln x)$, find ${\tfrac {dy}{dx}}$. Note: Another notation for $\sin ^{-1}$ is $\arcsin$

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
This is a composition of functions. What rule helps differentiate compositions of functions?

Hint 2
Use the chain rule, namely ${\tfrac {d}{dx}}f(g(x))=f'(g(x))g'(x)$

Hint 3
A reminder that ${\tfrac {d}{dx}}\sin ^{-1}(x)={\frac {1}{\sqrt {1-x^{2}}}}$

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. Since ${\tfrac {d}{dx}}\sin ^{-1}(x)={\frac {1}{\sqrt {1-x^{2}}}}$ and ${\tfrac {d}{dx}}\ln(x)={\frac {1}{x}}$ The chain rule gives us ${\tfrac {d}{dx}}\sin ^{-1}(\ln(x))={\frac {1}{\sqrt {1-(\ln(x))^{2}}}}\cdot {\frac {1}{x}}$

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. For an alternative solution, taking the sin of both sides yields $\displaystyle \sin(y)=\ln(x)$ Taking implicit derivatives of both sides yields $\displaystyle \cos(y)y'={\frac {1}{x}}$ Solving for the derivative yields $\displaystyle y'={\frac {1}{\cos(y)x}}$ Next, we want to express the right hand side as a function of x. Since $\displaystyle \sin(y)=\ln(x)$ and $\displaystyle \sin(y)$ is opposite over hypotenuse, we can draw the following triangle, interpreting y as an angle: where the last side (the base of the triangle) is computed using the Pythagorean theorem. Thus we see that $\displaystyle \cos(y)={\sqrt {1-(\ln(x))^{2}}}$ and plugging into the derivative yields $\displaystyle y'={\frac {1}{x{\sqrt {1-(\ln(x))^{2}}}}}$ completing the problem.

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