Science:Math Exam Resources/Courses/MATH104/December 2012/Question 01 (n)/Solution 1

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 $f$:

${\color {blue}f'(x)={\frac {1}{1+\ln(1+\ln(x))}}\cdot {\frac {1}{1+\ln(x)}}\cdot {\frac {1}{x}}}.$