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 $\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( \log x \right)=-{{x}^{-2}}$.