Question

Find the second derivative of the following function.
\[y=\arcsin (\ln x)\]

Answer 1

Hint: Here ‘arc’ is used for representing the inverse form of any trigonometric function. So, $\arcsin \left( \ln x \right)={{\sin }^{-1}}\left( \ln x \right)$. Apply chain rule for differentiating the given function which is given as $f\left( g\left( x \right) \right)'=f'\left( g\left( x \right) \right)g'\left( x \right)$.
Use $\dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$and $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$to simplify it further.

Complete step by step answer:
Here, we have given equation
\[y=\arcsin (\ln x)\]
Or \[y={{\sin }^{-1}}\left( \ln x \right)-(1)\]
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{\sin }^{-1}}\left( \ln x \right) \right) \\
 & \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{\left( \ln x \right)}^{2}}}}\dfrac{d}{dx}\left( \ln x \right)\left( \because \dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right) \\
 & \dfrac{dy}{dx}=\dfrac{1}{x\sqrt{1-{{\left( \ln x \right)}^{2}}}}\left( \because \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} \right) \\
\end{align}\]
Now, we have \[\dfrac{dy}{dx}\]as
\[\dfrac{dy}{dx}=\dfrac{1}{x\sqrt{1-{{\left( \ln x \right)}^{2}}}}-(2)\]
Now, we have to find the derivative of equation (2) as we need to calculate the second derivative of given function y.
Hence, we can write the equation (2) as
\[\dfrac{dy}{dx}={{x}^{-1}}{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}\]
Now, let us differentiate \[\dfrac{dy}{dx}\]again: -
\[\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( {{x}^{-1}}{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}} \right)-(3)\]
Now, from the equation (3) we can observe that there are two functions in multiplication, \[{{x}^{-1}}\]and \[{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}\].
Hence, here we need to apply multiplication rule of differentiation in a following way: -
Let us assume two functions u(x) and v(x) so that they are in multiplication form as follows: -
\[\dfrac{d}{dx}\left( u\left( x \right).v\left( x \right) \right)=u\left( x \right)\dfrac{d}{dx}\left( v\left( x \right) \right)+v\left( x \right)\dfrac{d}{dx}\left( u\left( x \right) \right)-(4)\]
Applying multiplication rule of differentiation in equation (3) as stated in equation (4) as follows: -
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( {{x}^{-1}}{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}} \right)\]
Here, from the equation (4)
\[u={{x}^{-1}},v={{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}\]
Therefore, we can write \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]as
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{x}^{-1}}\dfrac{d}{dx}{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}+{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}\dfrac{d}{dx}\left( {{x}^{-1}} \right)\]
As we can see that we need to apply chain rule as well wherever necessary because \[{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}\]is implicit function as stated in the beginning of solution. Let us simply \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]more: -
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{x}^{-1}}\times \left( \dfrac{-1}{2} \right){{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}-1}}.\dfrac{d}{dx}\left( 1-{{\left( \ln x \right)}^{2}} \right)+{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}\times \left( -1 \right){{x}^{-2}}\left( \because \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right) \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-1}{2x}{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-3}{2}}}\left( -2\ln x\times \dfrac{d}{dx}\left( \ln x \right) \right)-\dfrac{{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}}{{{x}^{2}}}\left( \because \dfrac{d}{dx}\ln x=\dfrac{1}{x} \right) \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-3}{2}}}\ln x}{2{{x}^{2}}}-\dfrac{{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}}{{{x}^{2}}} \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\ln x{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-3}{2}}}-{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}}{{{x}^{2}}} \\
\end{align}\]
Or
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\ln x}{{{x}^{2}}\left( 1-{{\left( \ln x \right)}^{2}} \right)\sqrt{1-{{\left( \ln x \right)}^{2}}}}-\dfrac{1}{{{x}^{2}}\sqrt{1-{{\left( \ln x \right)}^{2}}}}\]
Note:
One needs to be careful while applying chain rule and multiplication rule of differentiation. While applying chain one needs to differentiate all the functions involved.
Students can also apply \[\left( \dfrac{u}{v} \right)\]rule of differentiation while differentiating \[\dfrac{d}{dx}\left( {{x}^{-1}}{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}} \right)\]as following way: -
Let us assume two functions u(x) and v(x); Now let us differentiate \[y=\dfrac{u\left( x \right)}{v\left( x \right)}\]
\[\dfrac{dy}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]
Now, coming to \[\dfrac{d}{dx}\left( {{x}^{-1}}{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}} \right)\]we can write the above expression as \[\dfrac{d}{dx}\left( \dfrac{{{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}}{x} \right)\]and we can take \[u={{\left( 1-{{\left( \ln x \right)}^{2}} \right)}^{\dfrac{-1}{2}}}\]and \[v=x\] apply division rule \[\left( \dfrac{u}{v} \right)\]as stated above and get the same answer.