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How do you find the second derivative of ${{\left( \ln x \right)}^{2}}$ ?

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Last updated date: 20th Jun 2024
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Answer
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Hint: To find the second derivative of the above function ${{\left( \ln x \right)}^{2}}$, we have to first find the first derivative of this function and then again take the derivative of this first derivative. The property which we will require to take the derivative is the chain rule.

Complete step-by-step solution:
The function given in the above problem which we have to take the derivative of is as follows:
${{\left( \ln x \right)}^{2}}$
Let us equate the above function to y we get,
\[\Rightarrow y={{\left( \ln x \right)}^{2}}\]
Now, we are going to take the derivative with respect to x on both the sides.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d{{\left( \ln x \right)}^{2}}}{dx}$
The derivative of the above function is taken by using the chain rule in which we are substituting $\ln x=t$ in the above equation. When substituting $\ln x=t$ in the above equation we have to take the derivative of this function with respect to t so for that we are going to make some additions in the above equation as follows:
$\Rightarrow \dfrac{dy}{dt}\left( \dfrac{dt}{dx} \right)$ …………(1)
Now, our y becomes ${{t}^{2}}$ so we have to take the derivative of y with respect to t then will multiply the reciprocal of $\dfrac{dx}{dt}$.
Now, we are going to take the differentiation of $\ln x=t$ with respect to t on both the sides we get,
$\begin{align}
  & \Rightarrow \ln x=t \\
 & \Rightarrow \dfrac{1}{x}\dfrac{dx}{dt}=1 \\
\end{align}$
Rearranging the above equation in such a way so that we will get the expression of $\dfrac{dt}{dx}$ is as follows:
$\Rightarrow \dfrac{dt}{dx}=\dfrac{1}{x}$ ……………….. (2)
Now, we are going to take the derivative of ${{t}^{2}}$ with respect to t.
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d\left( {{t}^{2}} \right)}{dt}$
We know the derivative of ${{x}^{n}}$ where n is a positive integer is as follows:
$\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$
Using the above derivative in calculating the differentiation of ${{t}^{2}}$ with respect to t as follows:
$\dfrac{dy}{dt}=\dfrac{d\left( {{t}^{2}} \right)}{dt}=2t$……………(3)
Substituting the value of (2) and (3) in (1) we get,
$\Rightarrow 2t\left( \dfrac{1}{x} \right)$
Now, substituting t as $\ln x$ in the above equation we get,
$\Rightarrow \dfrac{dy}{dx}=2\ln x\left( \dfrac{1}{x} \right)$
Now, we are going to take the second derivative of the above first derivative. As you can see that the two functions $\ln x\And x$ are in division with respect to each other so we are going to use quotient rules to find the derivative of.
We know the quotient rule of two functions ${{f}_{1}}\left( x \right)\And {{f}_{2}}\left( x \right)$ is as follows:
$\dfrac{d\left( \dfrac{{{f}_{1}}\left( x \right)}{{{f}_{2}}\left( x \right)} \right)}{dx}=\dfrac{{{f}_{2}}\left( x \right){{f}_{1}}'\left( x \right)-{{f}_{1}}\left( x \right){{f}_{2}}'\left( x \right)}{{{\left( {{f}_{2}}\left( x \right) \right)}^{2}}}$
The ‘ in the above functions ${{f}_{1}}\And {{f}_{2}}$ is showing the first derivative of those functions with respect to x.
Now, in the above quotient rule, substituting the following values of ${{f}_{1}}\And {{f}_{2}}$ we get,
$\begin{align}
  & {{f}_{1}}\left( x \right)=2\ln x, \\
 & {{f}_{2}}\left( x \right)=x \\
\end{align}$
$\begin{align}
  & \dfrac{d\left( \dfrac{2\ln x}{x} \right)}{dx}=\dfrac{\left( x \right)\left( 2\ln x \right)'-\left( 2\ln x \right)\left( x \right)'}{{{\left( x \right)}^{2}}} \\
 & \Rightarrow \dfrac{d\left( \dfrac{2\ln x}{x} \right)}{dx}=\dfrac{x\left( \dfrac{2}{x} \right)-2\ln x\left( 1 \right)}{{{\left( x \right)}^{2}}} \\
\end{align}$
In the numerator of the above equation x will be cancelled out from the numerator and the denominator and we get,
$\begin{align}
  & \Rightarrow \dfrac{d\left( \dfrac{2\ln x}{x} \right)}{dx}=\dfrac{2-2\ln x\left( 1 \right)}{{{\left( x \right)}^{2}}} \\
 & \Rightarrow \dfrac{d\left( \dfrac{2\ln x}{x} \right)}{dx}=\dfrac{2\left( 1-\ln x \right)}{{{\left( x \right)}^{2}}} \\
\end{align}$
Hence, the second derivative of the above function is equal to $\dfrac{2\left( 1-\ln x \right)}{{{\left( x \right)}^{2}}}$.

Note: The mistake that could be possible in the above problem is that when we were finding the first derivative of the given function and we have applied the chain rule then we might forget to take the derivative of $\ln x$ with respect to x. And the first derivative will look like:
$\dfrac{dy}{dx}=2\ln x$
In the above derivative, we forgot to multiply the derivative of $\ln x$ so don’t make this mistake while solving such derivatives.