Question

Find the second derivative of the following function.
\[y={{\ln }^{2}}x-(\ln (\ln x))\]

Answer 1

Hint: Use Chain rule of differentiation which is given as
$\left( f\left( g\left( x \right) \right) \right)'=f'\left( g\left( x \right) \right)g'\left( x \right)$
Apply multiplication rule of differentiation or division rule whenever required.
Multiplication Rule:
$\dfrac{d}{dx}\left( u.v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$
Division Rule:
$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$
Use $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$.

Complete step by step answer:
We have given equation
\[y={{\ln }^{2}}x-(\ln (\ln x))\]
Let us differentiate the given equation w.r.t x
\[\dfrac{dy}{dx}=\dfrac{d}{dx}({{\ln }^{2}}x-\ln (\ln x))\]
Here we need to apply the chain rule of differentiation with \[{{\ln }^{2}}x\]and \[\ln (\ln x)\]as both are a combination of two functions. Now let us first see the chain rule of differentiation: -
If we have two functions and get combined as \[f\left( g\left( x \right) \right)\]and differentiation of \[f\left( g\left( x \right) \right)\]is calculated in following way: -
  \[f{{(g(x))}^{'}}={{f}^{'}}g(x){{g}^{'}}(x)-(2)\]
Now, coming to the question part: -
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{\ln }^{2}}x-\ln \left( \ln x \right) \right)=\dfrac{d}{dx}\left( {{\ln }^{2}}x \right)-\dfrac{d}{dx}\left( \ln \left( \ln x \right) \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{\ln }^{2}}x \right)-\dfrac{d}{dx}\left( \ln \left( \ln x \right) \right) \\
 & \Rightarrow \dfrac{dy}{dx}=2{{\left( \ln x \right)}^{1}}\dfrac{d}{dx}\left( \ln x \right)-\dfrac{1}{\ln x}\dfrac{d}{dx}\left( \ln x \right)\left( \because \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x},\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{2\ln x}{x}-\dfrac{1}{x\left( \ln x \right)}-(3) \\
\end{align}\]
Now, as from the question we need to calculate the second derivative of the given function. Therefore,
\[\begin{align}
  & \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{2\ln x}{x}-\dfrac{1}{x\left( \ln x \right)} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\dfrac{d}{dx}\left( \dfrac{\ln x}{x} \right)-\dfrac{d}{dx}\left( {{x}^{\left( -1 \right)}}{{\left( \ln x \right)}^{\left( -1 \right)}} \right)-(4) \\
\end{align}\]
Let us calculate \[\dfrac{d}{dx}\left( \dfrac{\ln x}{x} \right)\]and \[\dfrac{d}{dx}\left( {{x}^{\left( -1 \right)}}{{\left( \ln x \right)}^{\left( -1 \right)}} \right)\]individually as follows: -
Let \[A\left( x \right)=\dfrac{\ln x}{x}\]or \[B\left( x \right)=\dfrac{1}{x\ln x}\]or \[{{x}^{-1}}{{\left( \ln x \right)}^{-1}}\]
Now, we have to calculate \[\dfrac{dA}{dx}\]and \[\dfrac{dB}{dx}\].
\[A=\dfrac{\ln x}{x}\]
Here we can see that \[\ln x\]and x are in division so we can apply division rule of differentiation as stated below: -
If u(x) and v(x) are two functions as \[y=\dfrac{u}{v}\]then differentiation of y is given as \[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]
Therefore, \[\dfrac{dA}{dx}=\dfrac{d}{dx}\left( \dfrac{\ln x}{x} \right)\]
Here, \[u=\ln x,v=x\]
Using division rule as stated above for calculation of \[\dfrac{dA}{dx}\]: -
\[\begin{align}
  & \dfrac{dA}{dx}=\dfrac{d}{dx}\left( \dfrac{\ln x}{x} \right)=\dfrac{x\dfrac{d}{dx}\left( \ln x \right)-\left( \ln x \right)\dfrac{d}{dx}\left( x \right)}{{{x}^{2}}} \\
 & \Rightarrow \dfrac{dA}{dx}=\dfrac{1-\ln x}{{{x}^{2}}}-(5)\left( \because \dfrac{d}{dx}\left( \ln x \right)=1,\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right) \\
\end{align}\]
Now let us calculate \[\dfrac{dB}{dx}\]or \[\dfrac{d}{dx}{{x}^{-1}}{{\left( \ln x \right)}^{-1}}\]. As we have two functions \[{{\left( x \right)}^{-1}}\]and \[{{\left( \ln x \right)}^{-1}}\]in multiplication so we need to apply multiplication rule of differentiation as stated below: -
If we have two functions u(x) and v(x) in multiplication like y=uv then we can write \[\dfrac{dy}{dx}\]as
\[\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
And now we have \[u={{x}^{-1}}\]and \[v={{\left( \ln x \right)}^{-1}}\]in \[B={{x}^{-1}}{{\left( \ln x \right)}^{-1}}\]therefore,
\[\begin{align}
  & \dfrac{dB}{dx}=\dfrac{d}{dx}\left( {{x}^{-1}}{{\left( \ln x \right)}^{-1}} \right)={{x}^{-1}}\dfrac{d}{dx}{{\left( \ln x \right)}^{-1}}+{{\left( \ln x \right)}^{-1}}\dfrac{d}{dx}\left( {{x}^{-1}} \right) \\
 & \Rightarrow \dfrac{dB}{dx}={{x}^{-1}}\times \left( -1 \right){{\left( \ln x \right)}^{-2}}\dfrac{d}{dx}\left( \ln x \right)+\dfrac{{{\left( \ln x \right)}^{-1}}\left( -1 \right)}{{{x}^{2}}} \\
\end{align}\]
(Chain rule applied with \[\dfrac{d}{dx}{{\left( \ln x \right)}^{-1}}\]as stated in equation (2) \[\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}\])
\[\dfrac{dB}{dx}=\dfrac{-1}{{{x}^{2}}{{\left( \ln x \right)}^{2}}}-\dfrac{1}{{{x}^{2}}\left( \ln x \right)}-(6)\]
Now, we have the values of \[\dfrac{dA}{dx}\]and \[\dfrac{dB}{dx}\]from equation (5) and equation (6) respectively. So, put these values in equation (4): -
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\left( \dfrac{1-\ln x}{{{x}^{2}}} \right)+\dfrac{1}{{{x}^{2}}{{\left( \ln x \right)}^{2}}}+\dfrac{1}{{{x}^{2}}\left( \ln x \right)}\]
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2\left( 1-\ln x \right){{\left( \ln x \right)}^{2}}+1+\ln x}{{{x}^{2}}{{\left( \ln x \right)}^{2}}}\]

Note:
Chain rule is the important part of this question. In the first step \[{{\ln }^{2}}x\]and \[\ln (\ln x)\]both are a combination of two functions. So, applying chain rule to them will give the correct solution. There is an another way of doing the differentiation of implicit functions as follows: -
Let \[y=\ln \left( \ln x \right)\]
Let \[\ln x=X\]
Now, \[y=\ln X\]
Differentiate it w.r.t x
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{1}{X}\dfrac{dX}{dx}\left( \because \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln x}\dfrac{d}{dx}\left( \ln x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\ln x}.\dfrac{1}{x}=\dfrac{1}{x\ln x} \\
\end{align}\]
Here, we can suppose any variable of the interior functions involved.
As we can get confused between uv and \[\left( \dfrac{v}{u} \right)\]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 the 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 (uv) rule.
Last important part of this question is calculation. And if we will 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.