Question

Find the second derivative of the following function.
\[y=\ln (\ln (\ln x))\]

Answer 1

Hint: We have to differentiate the function twice to get the second derivative. As the given function is a composite function so we will apply chain rule, i.e., \[f{{\left( g\left( x \right) \right)}^{'}}={{f}^{'}}\left( g\left( x \right) \right){{g}^{'}}\left( x \right)\] and use the formula \[\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}\]. After getting the first derivative, differentiate again using product and division rule \[\left( \dfrac{d}{dx}\left( u.v \right) \right)\] and \[\left( \dfrac{d}{dx}\left( \dfrac{u}{v} \right) \right)\] to get the second derivative.

Complete step by step answer:
 We have the function as \[y=\ln (\ln (\ln x))-(1)\]
Here we can observe that the given function is a combination of three functions, hence it can be termed as a composite function. So, here we need to apply the chain rule of differentiation as stated below: -
If we have a composite function as \[f\left( g\left( x \right) \right)\]then derivative of this function can be written as \[f{{\left( g\left( x \right) \right)}^{'}}={{f}^{'}}\left( g\left( x \right) \right){{g}^{'}}\left( x \right)-(2)\]
It means we need to differentiate a composite function one by one in a continuous manner.
Let us know differentiate the first equation of given function by using the chain rule defined in equation (2): -
\[\begin{align}
  &\Rightarrow y=\ln \left( \ln \left( \ln x \right) \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \ln \left( \ln \left( \ln x \right) \right) \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln \left( \ln x \right)}\dfrac{d}{dx}\left( \ln \left( \ln x \right) \right)\left( \because \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln \left( \ln x \right)}.\dfrac{1}{\ln x}\dfrac{d}{dx}\left( \ln x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln \left( \ln x \right)}.\dfrac{1}{\ln x}.\dfrac{1}{x}\left( \because \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} \right) \\
\end{align}\]
Therefore, we have the equation
 \[\dfrac{dy}{dx}=\dfrac{1}{x\left( \ln x \right)\ln \left( \ln x \right)}-(3)\]
Now, coming to the question, as we need to calculate the second derivative of the given function i.e. \[\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)\] or \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].
Hence, let us differentiate the equation (3) again to get \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].
\[\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \left( {{x}^{-1}} \right){{\left( \ln x \right)}^{-1}}{{\left( \ln \left( \ln x \right) \right)}^{-1}} \right)\]
Here, we can observe that we have the functions in multiplication as \[\dfrac{1}{x},\dfrac{1}{\ln x},\dfrac{1}{\ln \left( \ln x \right)}\]. We already know the formula of \[\dfrac{d}{dx}\left( u.v \right)\] and \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)\] i.e. two functions in multiplication and division respectively as stated below: -
\[\begin{align}
  & \dfrac{d}{dx}\left( u.v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}-(4) \\
 & \dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}+u\dfrac{dv}{dx}}{{{v}^{2}}}-(5) \\
\end{align}\]
Now, here we have three functions in multiply, so let us first derive the formula for it.
Let us assume three functions as u, v and w. So, we need to find \[\dfrac{d}{dx}\left( u.v.w \right)=?\]
Let us suppose u.v=X and then apply multiplication rule or \[\dfrac{d}{dx}\left( u.v \right)\] rule defined in equation (4) as follows: -
\[\dfrac{d}{dx}\left( Xw \right)=X\dfrac{dw}{dx}+w\dfrac{dX}{dx}\]
Now, we have X=u.v, so putting the values of X, we get
\[\dfrac{d}{dx}\left( u.v.w \right)=u.v\dfrac{dw}{dx}+w\dfrac{d}{dx}\left( u.v \right)\]
Again, applying formula for
\[\dfrac{d}{dx}\left( u.v \right)\] from equation (4) as follows: -
\[\begin{align}
  & \dfrac{d}{dx}\left( u.v.w \right)=u.v\dfrac{dw}{dx}+w\left( v\dfrac{dv}{dx}+v\dfrac{du}{dx} \right) \\
 & \dfrac{d}{dx}\left( u.v.w \right)=u.v\dfrac{dw}{dx}+uw\dfrac{dv}{dx}+vw\dfrac{du}{dx}-(6) \\
\end{align}\]
Now, let us use the derived formula (6) in equation (3) by assuming
\[\begin{align}
  & u={{\left( x \right)}^{-1}} \\
 & v={{\left( \ln x \right)}^{-1}} \\
 & w={{\left( \ln \left( \ln x \right) \right)}^{-1}} \\
\end{align}\]
Hence, putting values in equation (6).
Let us calculate \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]: -
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{x\ln x}\dfrac{d}{dx}{{\left( \ln \left( \ln x \right) \right)}^{-1}}+\dfrac{1}{x\left( \ln \left( \ln x \right) \right)}\dfrac{d}{dx}{{\left( \ln x \right)}^{-1}}+\dfrac{1}{\ln x.\ln \left( \ln x \right)}\dfrac{d}{dx}\left( {{x}^{-1}} \right)\]
Now, here we need to apply chain rule of differentiation as well to get the correct results as stated in equation (2).
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{x\ln x}\left( -1 \right){{\left( \ln \left( \ln x \right) \right)}^{-2}}\dfrac{d}{dx}\left( \ln \left( \ln x \right) \right)+\dfrac{1}{x\ln \left( \ln x \right)}\left( -1 \right){{\left( \ln x \right)}^{-2}}\dfrac{d}{dx}\left( \ln x \right)+\dfrac{1}{\ln x\ln \left( \ln x \right)}\left( -1 \right){{\left( x \right)}^{-2}}\dfrac{d}{dx}\left( x \right)\left( \because \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{x\left( \ln x \right){{\left( \ln \left( \ln x \right) \right)}^{2}}}\dfrac{1}{\ln x}.\dfrac{d}{dx}\left( \ln x \right)+\dfrac{-1}{x\ln \left( \ln x \right){{\left( \ln x \right)}^{2}}}\times \dfrac{1}{x}+\dfrac{-1}{{{x}^{2}}\ln x\ln \left( \ln x \right)}\left( \because \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{x{{\left( \ln x \right)}^{2}}{{\left( \ln \left( \ln x \right) \right)}^{2}}}-\dfrac{1}{{{x}^{2}}{{\left( \ln x \right)}^{2}}\ln \left( \ln x \right)}-\dfrac{1}{{{x}^{2}}\left( \ln x \right)\ln \left( \ln x \right)} \\
\end{align}\]
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1-\ln \left( \ln x \right)-\left( \ln x \right)\left( \ln \left( \ln x \right) \right)}{{{x}^{2}}{{\left( \ln x \right)}^{2}}{{\left( \ln \left( \ln x \right) \right)}^{2}}}\]

Note:
Chain rule is the important part of this question. In the first step \[{{\left( \ln x \right)}^{2}}\] and \[\ln \left( \ln x \right)\] both are combinations of two functions. So, applying chain rule to them will give the correct solution. There is another way of doing the differentiation of composite functions as follows: -
\[y=\ln \left( \ln \left( \ln x \right) \right)\]
Let, \[\ln \left( \ln x \right)=X\]
Now, differentiating it w.r.t x
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{1}{X}\dfrac{dX}{dx}\left( \because \dfrac{d}{dx}\ln x=\dfrac{1}{x} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln \left( \ln x \right)}.\dfrac{d}{dx}\left( \ln \left( \ln x \right) \right) \\
\end{align}\]
Put \[\left( \ln x \right)=K\]
Hence, we can write \[\dfrac{dy}{dx}\] as
\[\begin{align}
  &\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln \left( \ln x \right)}\dfrac{d}{dx}\ln K \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln \left( \ln x \right)}.\dfrac{1}{K}\dfrac{dK}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln \left( \ln x \right)\left( \ln x \right)}.\dfrac{d}{dx}\left( \ln x \right)\left( \because K=\ln x,\dfrac{d}{dx}\ln x=\dfrac{1}{x} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x\ln \left( \ln x \right)\left( \ln x \right)} \\
\end{align}\]
As we can get confused between u.v and \[\dfrac{v}{u}\] rule’s application. Students need to know very clearly that both can be applied anywhere with \[\left( \dfrac{\sin x}{x} \right)\] or \[\left( \sin x.\ln x \right)\] both.
Let we have function \[\left( \dfrac{\sin x}{x} \right)\] then we can observe that \[\left( \dfrac{u}{v} \right)\] rule should be applicable as \[u=\sin x\] and \[v=x\] but if we want to apply (u.v) rule then we need to consider \[u=\sin x\] and \[v=\dfrac{1}{x}\] or \[{{x}^{-1}}\]. Hence, we need not be confused anywhere between \[\left( \dfrac{u}{v} \right)\] and \[\left( uv \right)\] rule.
Last important part of this question is calculation. And if we will not write even a single wrong term during calculation it can lead us to the wrong answer and for that we need to check for the mistake from the initial level. Hence, calculations play an important role in these kinds of questions.