Question

Find the second derivative of function \[\log x\].

Answer 1

Hint: To find the derivative of a function we should know what the first principle of the derivative is and how to apply it. First principle of derivative states that, \[{{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{x+h-x}\]. For \[\log x\], keep in mind that you will require the expansion of \[\log \left( 1+x \right)\] which is \[\log \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+.....\].With help of these concepts we try to solve the question.

Complete step by step answer:
In this question, we have to find the second derivative of \[\log x\], for that, we should know what is the first derivative of the function \[\log x\]. To find that, we will apply first principle of derivatives, which states that, \[{{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{x+h-x}\].
We have been given that \[f\left( x \right)=\log x\], which means,
\[{{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\log \left( x+h \right)-\log \left( x \right)}{x+h-x}......\left( i \right)\]
We know that, \[\log a-\log b=\log \left( \dfrac{a}{b} \right)\]. Therefore, we can write equation (i) can be written as,
\[{{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\log \left( \dfrac{x+h}{x} \right)}{x+h-x}\], which can be written as
\[\Rightarrow {{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+\dfrac{h}{x} \right)}{x+h-x}......\left( ii \right)\]
As we know the expansion of \[\log \left( 1+x \right)\], which is \[\log \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+.....\], we are using this value in equation (ii), we will get,
\[{{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\left( \dfrac{h}{x} \right)-\dfrac{{{\left( \dfrac{h}{x} \right)}^{2}}}{2}+\dfrac{{{\left( \dfrac{h}{x} \right)}^{3}}}{3}-\dfrac{{{\left( \dfrac{h}{x} \right)}^{4}}}{4}+.....}{x+h-x}\]
Now, we will take \[\left( \dfrac{h}{x} \right)\] common from numerator, so we will get,
\[{{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{h}{x} \right)\dfrac{1-\dfrac{{{\left( \dfrac{h}{x} \right)}^{1}}}{2}+\dfrac{{{\left( \dfrac{h}{x} \right)}^{2}}}{3}-\dfrac{{{\left( \dfrac{h}{x} \right)}^{3}}}{4}+.....}{h}\]
In this, we can see that ‘h’ can be cancelled out from numerator and denominator, so we will get,
\[{{f}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1-\dfrac{{{\left( \dfrac{h}{x} \right)}^{1}}}{2}+\dfrac{{{\left( \dfrac{h}{x} \right)}^{2}}}{3}-\dfrac{{{\left( \dfrac{h}{x} \right)}^{3}}}{4}+.....}{x}\]
Now, we are putting the limits, so we will get,
\[{{f}^{'}}\left( x \right)=\dfrac{1-\dfrac{{{\left( \dfrac{0}{x} \right)}^{1}}}{2}+\dfrac{{{\left( \dfrac{0}{x} \right)}^{2}}}{3}-\dfrac{{{\left( \dfrac{0}{x} \right)}^{3}}}{4}+.....}{x}\]
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{1-0}{x}\]
\[\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{1}{x}\]
Now, we can say that the first derivative of \[\log x\] is \[\dfrac{1}{x}\].
Now, to find the second derivative, we will apply the first principle of derivatives on \[\dfrac{1}{x}\], as it is the first derivative of \[\log x\].
Let us consider, \[g\left( x \right)=\dfrac{1}{x}\]. Therefore, we can say that, \[{{g}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{x+h-x}\]
Now, we will simplify above equation further, so we get,
\[\Rightarrow {{g}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}\]
\[\Rightarrow {{g}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{x-x-h}{\left( x+h \right)\left( x \right)}}{h}\]
\[\Rightarrow {{g}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{-h}{\left( x+h \right)\left( x \right)}}{h}\]
\[\Rightarrow {{g}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-h}{h\left( x+h \right)\left( x \right)}\]
\[\Rightarrow {{g}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-1}{\left( x+h \right)\left( x \right)}\]
Now, we are putting limits, so we get,
\[\Rightarrow {{g}^{'}}\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{-1}{\left( x+0 \right)\left( x \right)}\]
\[\Rightarrow {{g}^{'}}\left( x \right)=\dfrac{-1}{\left( x \right)\left( x \right)}\]
\[\Rightarrow {{g}^{'}}\left( x \right)=\dfrac{-1}{{{\left( x \right)}^{2}}}\]
\[\Rightarrow {{f}^{''}}\left( x \right)=\dfrac{-1}{{{\left( x \right)}^{2}}}\]
Therefore, we can say that the second derivative of \[\log x\] is \[\dfrac{-1}{{{\left( x \right)}^{2}}}\].

Note: In this question, the most important thing to remember is the expansion of \[\log \left( 1+x \right)\], which is \[\log \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+.....\], because if we will not apply this expansion our first derivative will become zero and so as second derivative which is wrong. Also, we can use a shortcut method for finding the second derivative of \[\log x\], that is, by remembering its first derivative, which is, \[\dfrac{1}{x}\]. Now, we can write this as \[{{x}^{-1}}\]and apply the formula of derivative of \[{{x}^{n}}\], which is equal to \[n{{x}^{n-1}}\]. Therefore, we can write, \[\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=-{{x}^{-2}}\], which is as same as \[\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( \log x \right)=-{{x}^{-2}}\].