Question

How do you find the second derivative of $\ln \left( {{x}^{\dfrac{1}{2}}} \right)$?

Answer 1

In the given question, we have been asked to find the second derivation of the given function. In order to solve the given question, first we need to derive the given function with respect to ‘x’ and simplify. Later we derivative again the resultant first derivation and simplify further, we will get our second derivative of $\ln \left( {{x}^{\dfrac{1}{2}}} \right)$.

Complete step by step solution:
We have given that,
$\ln \left( {{x}^{\dfrac{1}{2}}} \right)$
Let $f\left( x \right)=\ln \left( {{x}^{\dfrac{1}{2}}} \right)$
Using the definition of law, i.e.
$\log \left( {{x}^{a}} \right)=a\log \left( x \right)$
Applying the rule in the given function, we get
$f\left( x \right)=\dfrac{1}{2}\ln \left( x \right)$
Derivative the above function with respect to ‘x’, we get
$f'\left( x \right)=\dfrac{d}{dx}\left( \dfrac{1}{2}\ln \left( x \right) \right)$
As we know that, $\dfrac{d}{dx}\left( k\times f\left( x \right) \right)=k\dfrac{d}{dx}f\left( x \right)$
Therefore,
$f'\left( x \right)=\dfrac{1}{2}\dfrac{d}{dx}\left( \ln \left( x \right) \right)$
Derivate ln(x) in the above function, we get
$f'\left( x \right)=\dfrac{1}{2}.\dfrac{1}{x}$
$f'\left( x \right)=\dfrac{1}{2}{{x}^{-1}}$
Now,
Derivation the resultant first derivative with respect to ‘x’, we get
We know that, $f''x=\dfrac{d}{dx}f'x$
We have,
$f'\left( x \right)=\dfrac{1}{2}{{x}^{-1}}$
Thus,
$f''\left( x \right)=\dfrac{d}{dx}\left( \dfrac{1}{2}{{x}^{-1}} \right)$
As we know that, $\dfrac{d}{dx}\left( k\times f\left( x \right) \right)=k\dfrac{d}{dx}f\left( x \right)$
Therefore,
$f''\left( x \right)=\dfrac{1}{2}\dfrac{d}{dx}\left( {{x}^{-1}} \right)$
As we know that the derivation of ${{x}^{n}}=n{{x}^{n-1}}$
Thus,
$f''\left( x \right)=\dfrac{1}{2}\left( -1\times {{x}^{-1-1}} \right)$
Simplifying the above, we get
$f''\left( x \right)=\dfrac{1}{2}\left( -{{x}^{-2}} \right)$
$f''\left( x \right)=-\dfrac{1}{2}\left( {{x}^{-2}} \right)$
$f''\left( x \right)=-\dfrac{1}{2{{x}^{2}}}$

Therefore the second derivative of $\ln \left( {{x}^{\dfrac{1}{2}}} \right)$ is $-\dfrac{1}{2{{x}^{2}}}$.

Note: A differential equation is the equation that contains function with one or more of its derivatives and differentials. The only mistakes that could be possible while doing the question i.e. finding the second derivation is that we might forget the formulas of differentiation and the way we differentiate a given function.