Question

Find the second derivative of the following function.
\[y=\sqrt[3]{\ln (\sin x)}\]

Answer 1

Hint: We have to differentiate the given function twice to get the second derivative. For getting the first derivative use the chain rule for differentiating the given function, which can be given as, $\left( f\left( g\left( x \right) \right) \right)'=f'\left( g\left( x \right) \right).g'\left( x \right)$ and also apply the formulas $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}},\dfrac{d}{dx}\sin x=\cos x,\dfrac{d}{dx}\ln x=\dfrac{1}{x},\dfrac{d}{dx}\cos x=-\sin x$. You will get First derivative, then differentiate again to using multiplication Rule and division Rule of differentiation to get the final answer.

Complete step by step answer:
We have the given equation/function as
\[y={{(\ln (\sin x))}^{1/3}}-(1)\]
Here, we need to apply chain rule for the differentiation of equation (1) as the given function is an implicit function i.e. combination of two or more functions.
As, we know the differentiation of implicit function is done in following way: -
\[\dfrac{d}{dx}f\left( g\left( x \right) \right)={{\left( f\left( g\left( x \right) \right) \right)}^{'}}={{f}^{'}}\left( g\left( x \right) \right){{g}^{'}}\left( x \right)-(2)\]
Applying the chain rule as expressed in equation with the differentiation of equation (1)
\[\begin{align}
  & y={{\left( \ln \left( \sin x \right) \right)}^{\dfrac{1}{3}}} \\
 & \dfrac{dy}{dx}=\dfrac{d}{dx}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{1}{3}}} \\
 & \dfrac{dy}{dx}=\dfrac{1}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{1}{3}-1}}\dfrac{d}{dx}\left( \ln \left( \sin x \right) \right)\left( \because \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right) \\
 & \dfrac{dy}{dx}=\dfrac{1}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\dfrac{1}{\sin x}\dfrac{d}{dx}\left( \sin x \right)\left( \because \dfrac{d}{dx}\ln x=\dfrac{1}{x} \right) \\
 & \dfrac{dy}{dx}=\dfrac{1}{3}\dfrac{{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}}{\sin x}\cos x\left( \because \dfrac{d}{dx}\sin x=\cos x \right) \\
 & \because \dfrac{dy}{dx}=\dfrac{1}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\cot x-(3) \\
\end{align}\]
As we have to calculate the second derivative of the given function ‘y’. Let us differentiate again equation (3) in following way: -
\[\begin{align}
  & \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{1}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\cot x \right) \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{3}\dfrac{d}{dx}\left( {{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\cot x \right)-(4) \\
\end{align}\]
Now, from the equation (4) we can observe there are two functions in multiplication; \[{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\]and \[\cot x\].So, here we need to apply rule of multiplication in differentiation as explained below: -
Let us assume two function u(x) and v(x) so that they are in multiplication form as follows: -
\[\begin{align}
  & y=u(x)v(x) \\
 & \dfrac{dy}{dx}=v(x)\dfrac{d}{dx}\left( u\left( x \right) \right)+u\left( x \right)\dfrac{d}{dx}\left( v\left( x \right) \right)-(5) \\
\end{align}\]
Now, applying the rule of multiplication explained in equation (5) with equation (4): -
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{3}\dfrac{d}{dx}\left( \ln {{\left( \sin x \right)}^{\dfrac{-2}{3}}}\cot x \right)\]
Here, \[u={{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}},v=\cot x\]
Therefore, we can proceed further as
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{3}\left( \cot x\dfrac{d}{dx}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}+{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\dfrac{d}{dx}\cot x \right)\]
Now, we can observe that \[{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\]is an implicit function so here we need to apply chain rule as explained in equation (2).
Hence, we can simplify \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] using the formula \[\left( \because \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right)\] as,
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{3}\cot x\times \dfrac{-2}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}-1}}\dfrac{d}{dx}\left( \ln \left( \sin x \right) \right)+\dfrac{1}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\left( -{{\csc }^{2}}x \right)\]
Now we know, \[\left( \dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x} \right)\], so above equation can be written as,
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2}{9}\cot x{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-5}{3}}}\dfrac{1}{\sin x}\dfrac{d}{dx}\left( \sin x \right)+\dfrac{1}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\left( -{{\csc }^{2}}x \right) \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2}{9}\dfrac{\cot x{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-5}{3}}}}{\sin x}\cos x-\dfrac{1}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}.{{\csc }^{2}}x \\
\end{align}\]
Hence,
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2}{9}{{\cot }^{2}}x{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-5}{3}}}-\dfrac{1}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}{{\csc }^{2}}x\]
As we know that \[y={{\left( \ln \left( \sin x \right) \right)}^{\dfrac{1}{3}}}\]from the equation (1)
Therefore, we can write \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]as
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2}{9}{{\cot }^{2}}x{{y}^{-5}}-\dfrac{1}{3}{{y}^{-2}}{{\csc }^{2}}x \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2}{9}\dfrac{{{\cot }^{2}}x}{{{y}^{5}}}-\dfrac{1}{3}\dfrac{{{\csc }^{2}}x}{{{y}^{2}}} \\
\end{align}\]
This is the required second derivative of the given equation.

Note:
Chain rule is the most important part of this question. One can go wrong while applying it in a following way: -
\[\dfrac{d}{dx}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{1}{3}}}=\dfrac{1}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\cos x\]
Where we have missed the differentiation of \[\ln \left( \sin x \right)\]. Therefore, we need to apply chain rule of differentiation very carefully.
Calculation is also an important side of these kinds of questions.
One can apply division rule of differentiation for the \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]after calculating \[\dfrac{dy}{dx}\]in the following way: -
\[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}-(6)\]
From the equation (3) in solution
\[\dfrac{d}{dx}\left( \dfrac{1}{3}{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{-2}{3}}}\cot x \right)=\dfrac{d}{dx}\left( \left( \dfrac{1}{3} \right)\dfrac{\cot x}{{{\left( \ln \sin x \right)}^{\dfrac{2}{3}}}} \right)=\dfrac{1}{3}\dfrac{d}{dx}\left( \dfrac{\cot x}{{{\left( \ln \left( \sin x \right) \right)}^{\dfrac{2}{3}}}} \right)\]
Now, we can suppose \[u=\cot x\]and \[v={{\left( \ln \left( \sin x \right) \right)}^{\dfrac{2}{3}}}\]
And apply the division rule of differentiation as stated in equation (6).