Question

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

Answer 1

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 $y={{\sin }^{-1}}\left( \ln x \right)-(1)$

& \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{aligned}$$ Now, we have $$\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{aligned} & \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{aligned}$$ Or $$\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.