Question

How do you find the second derivative of $\ln \left( {{x}^{3}} \right)$?

Answer 1

The differentiation of the given function $\ln \left( {{x}^{3}} \right)$ will be defined as ${{f}^{'}}\left( x \right)$ and ${{f}^{''}}\left( x \right)$ respectively where $y=f\left( x \right)=\ln \left( {{x}^{3}} \right)$. We differentiate the given function $y=\ln \left( {{x}^{3}} \right)$ with $x$. The differentiated forms are $\dfrac{d}{dx}\left[ \ln \left( x \right) \right]=\dfrac{1}{x}$ and $\dfrac{d}{dx}\left[ {{x}^{n}} \right]=n{{x}^{n-1}}$. We use those forms and keep the constants as it is.

Complete step-by-step solution:
In case of finding any derivative, we have to differentiate the given function $y=\ln \left( {{x}^{3}} \right)$ with $x$. Let’s assume that $y=f\left( x \right)=\ln \left( {{x}^{3}} \right)$. The given function is a function of $x$.
The first derivative is $\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ f\left( x \right) \right]$. It’s also defined as ${{f}^{'}}\left( x \right)$.
The differentiated form of $\ln \left( x \right)$ is $\dfrac{d}{dx}\left[ \ln \left( x \right) \right]=\dfrac{1}{x}$. We use the chain rule for the composite function. Differentiating $f\left( x \right)=goh\left( x \right)$, we get
$\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ goh\left( x \right) \right]=\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}={{g}^{'}}\left[ h\left( x \right) \right]{{h}^{'}}\left( x \right)$.
We apply the chain rule for the composite function where the main function is $g\left( x \right)=\ln x$ and the other function is $h\left( x \right)={{x}^{3}}$.
Therefore, the first derivative of $y=\ln \left( {{x}^{3}} \right)$ is ${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left[ \ln \left( {{x}^{3}} \right) \right]=\dfrac{3{{x}^{2}}}{{{x}^{3}}}=\dfrac{3}{x}$.
Now we need to find the second derivative.
The second derivative is $\dfrac{d}{dx}\left[ {{f}^{'}}\left( x \right) \right]$. It’s also defined as ${{f}^{''}}\left( x \right)$.
There is a function $a f\left( x \right)$ where $a$ is a constant. If we are going to differentiate the function the formula remains as $\dfrac{d}{dx}\left[ af\left( x \right) \right]=a\dfrac{d}{dx}\left[ f\left( x \right) \right]$.
The differentiated form of $\dfrac{1}{x}$ is $\dfrac{d}{dx}\left[ \dfrac{1}{x} \right]=\dfrac{-1}{{{x}^{2}}}$. The constant remains as it is.
Therefore, the second derivative of ${{f}^{'}}\left( x \right)=\dfrac{3}{x}$ is ${{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left[ \dfrac{3}{x} \right]=\dfrac{-3}{{{x}^{2}}}$.
The second derivative of the given function $\ln \left( {{x}^{3}} \right)$ is $\dfrac{-3}{{{x}^{2}}}$ respectively.

Note: The second derivative can also be expressed as $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)$. The differentiation $\dfrac{d}{dx}\left[ \dfrac{1}{x} \right]=\dfrac{-1}{{{x}^{2}}}$ has been done following the rule of exponent where $\dfrac{d}{dx}\left[ {{x}^{n}} \right]=n{{x}^{n-1}}$. Here the value of n was -1 as ${{x}^{-1}}=\dfrac{1}{x}$. We need remember that in the chain rule $\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}$, we aren’t cancelling out the part $d\left[ h\left( x \right) \right]$. Cancelation of the base differentiation is never possible. It’s just a notation to understand the function which is used as a base to differentiate.