Science:Math Exam Resources/Courses/MATH104/December 2012/Question 01 (n)

MATH104 December 2012

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Question 01 (n)
Find the derivative of the function $\displaystyle {}f(x)=\ln(1+\ln(1+\ln(x))).$

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
We need the chain rule for this problem and might need to use it more than once. Can you identify an inner and outer function for $f(x)$?

Checking a solution serves two purposes: helping you if, after having used the hint, 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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. This question requires multiple uses of the chain rule. $\displaystyle f(x)=\ln(1+\ln(1+\ln(x)))$ Recognizing that the inner function is $u=h(x)=1+\ln(1+\ln(x))$ and the outer function is $g(u)=\ln(u)$, we apply the chain rule and get ${\begin{aligned}f'(x)&=g'(h(x))\cdot h'(x)\\&={\frac {1}{1+\ln(1+\ln(x))}}\cdot h'(x).\end{aligned}}$ Now we need to use the chain rule again to evaluate $h'(x)$. The function $h(x)$ has inner function $v=k(x)=1+\ln(x)$ and outer function $j(v)=1+\ln(v)$. Thus, $h'(x)$ is given by ${\begin{aligned}h'(x)&=j'(k(x))\cdot k'(x)\\&={\frac {1}{1+\ln(x)}}\cdot k'(x)\\&={\frac {1}{1+\ln(x)}}\cdot {\frac {1}{x}}\end{aligned}}$ Thus putting the pieces together, we write the expression for the derivative of $f$: ${\color {blue}f'(x)={\frac {1}{1+\ln(1+\ln(x))}}\cdot {\frac {1}{1+\ln(x)}}\cdot {\frac {1}{x}}}.$

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