Consider the given expression, \[y={{\sin }^{-1}}(a\sin x)\], find the value of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]

Hint: First modify the given expression in simplest form and then find the first and second order derivative. Now convert the second order derivative in terms of first order derivative.

Complete step-by-step answer:
The given expression is,
\[y={{\sin }^{-1}}(a \sin x)\]
We will directly apply the differentiation with respect to $'x'$ , we get
\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{\sin }^{-1}}(a\sin x) \right)\]
Now we will apply chain rule of differentiation and, \[\dfrac{d}{dx}({{\sin }^{-1}}x)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\], the above equation can be written as,
\[\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{(a\sin x)}^{2}}}}\dfrac{d}{dx}\left( a\sin x \right)\]
Now taking out the constant term from the differentiation, we get
\[\dfrac{dy}{dx}=\dfrac{a}{\sqrt{1-{{(a\sin x)}^{2}}}}\dfrac{d}{dx}\left( \sin x \right)\]
We know, \[\dfrac{d}{dx}(\sin x)=\cos x\], applying this formula the above equation can be written as,
\[\dfrac{dy}{dx}=\dfrac{a}{\sqrt{1-{{(a\sin x)}^{2}}}}(\cos x)\]
This is the first derivative. Let us differentiate again to find the second order derivative, so differentiating the above expression with respect to $'x'$, we get
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=a\dfrac{d}{dx}\left( \dfrac{\cos x}{\sqrt{1-{{(a\sin x)}^{2}}}} \right)\]
Applying the quotient rule, i.e., $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}(u)-u\dfrac{d}{dx}(v)}{{{v}^{2}}}$, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{\sqrt{1-{{(a\sin x)}^{2}}}\dfrac{d}{dx}(\cos x)-\cos x\dfrac{d}{dx}\left( \sqrt{1-{{(a\sin x)}^{2}}} \right)}{{{\left( \sqrt{1-{{(a\sin x)}^{2}}} \right)}^{2}}} \right)\]
We know, \[\dfrac{d}{dx}(cosx)=-\sin x\], applying this formula the above equation can be written as,
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{\sqrt{1-{{(a\sin x)}^{2}}}(-\sin x)-\cos x\dfrac{d}{dx}\left( \sqrt{1-{{(a\sin x)}^{2}}} \right)}{1-{{(a\sin x)}^{2}}} \right)\]
Now we know $\dfrac{d}{dx}({{u}^{n}})=n{{u}^{n-1}}\dfrac{d}{dx}(u)$ , applying this formula, the above equation becomes,
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{\sqrt{1-{{(a\sin x)}^{2}}}(-\sin x)-\cos x\left( \dfrac{1}{2}{{\left( 1-{{(a\sin x)}^{2}} \right)}^{\dfrac{1}{2}-1}} \right)\dfrac{d}{dx}\left( 1-{{(a\sin x)}^{2}} \right)}{1-{{(a\sin x)}^{2}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{\sqrt{1-{{(a\sin x)}^{2}}}(-\sin x)-\cos x\left( \dfrac{1}{2}{{\left( 1-{{(a\sin x)}^{2}} \right)}^{\dfrac{-1}{2}}} \right)\left( -{{a}^{2}}\dfrac{d}{dx}\left( {{\sin }^{2}}x \right) \right)}{1-{{(a\sin x)}^{2}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{\sqrt{1-{{(a\sin x)}^{2}}}(-\sin x)-\cos x\left( \dfrac{1}{2}{{\left( 1-{{(a\sin x)}^{2}} \right)}^{\dfrac{-1}{2}}} \right)\left( -{{a}^{2}}2\sin x\dfrac{d}{dx}\left( \sin x \right) \right)}{1-{{(a\sin x)}^{2}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{\sqrt{1-{{(a\sin x)}^{2}}}(-\sin x)-\cos x\left( \dfrac{1}{2}{{\left( 1-{{(a\sin x)}^{2}} \right)}^{\dfrac{-1}{2}}} \right)\left( -{{a}^{2}}2\sin x\cos x \right)}{1-{{(a\sin x)}^{2}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{\sqrt{1-{{(a\sin x)}^{2}}}(-\sin x)+\dfrac{\cos x\left( {{a}^{2}}2\sin x\cos x \right)}{2\sqrt{1-{{(a\sin x)}^{2}}}}}{1-{{(a\sin x)}^{2}}} \right) \\
\end{align}\]
Cancelling the like terms, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{\sqrt{1-{{(a\sin x)}^{2}}}(-\sin x)+\dfrac{{{a}^{2}}{{\cos }^{2}}x\sin x}{\sqrt{1-{{(a\sin x)}^{2}}}}}{1-{{(a\sin x)}^{2}}} \right)\]
Separating the terms, we get
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{-\sin x\sqrt{1-{{(a\sin x)}^{2}}}}{1-{{(a\sin x)}^{2}}}+\dfrac{\dfrac{{{a}^{2}}{{\cos }^{2}}x\sin x}{\sqrt{1-{{(a\sin x)}^{2}}}}}{1-{{(a\sin x)}^{2}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{-\sin x\sqrt{1-{{(a\sin x)}^{2}}}}{1-{{(a\sin x)}^{2}}}\times \dfrac{\sqrt{1-{{(a\sin x)}^{2}}}}{\sqrt{1-{{(a\sin x)}^{2}}}}+\dfrac{{{a}^{2}}{{\cos }^{2}}x\sin x}{{{\left( 1-{{(a\sin x)}^{2}} \right)}^{\dfrac{3}{2}}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\left( \dfrac{-\sin x\left( 1-{{(a\sin x)}^{2}} \right)}{\left( 1-{{(a\sin x)}^{2}} \right)\sqrt{1-{{(a\sin x)}^{2}}}}+\dfrac{{{a}^{2}}{{\cos }^{2}}x\sin x}{{{\left( 1-{{(a\sin x)}^{2}} \right)}^{\dfrac{3}{2}}}} \right) \\
\end{align}\]
Cancelling the like terms, we get
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{{{a}^{3}}{{\cos }^{2}}x\sin x}{{{\left( 1-{{(a\sin x)}^{2}} \right)}^{\dfrac{3}{2}}}}-\dfrac{a\sin x}{\sqrt{1-{{(a\sin x)}^{2}}}} \right)\]
This is the required second order derivative.

Note: Another approach of this problem is converting the given expression in simplified form, i.e.,
\[y={{\sin }^{-1}}(a \sin x)\]
Multiply both sides with $\sin $ , we get
\[\sin y=\sin ({{\sin }^{-1}}(a\sin x))\]
Now $\sin ,{{\sin }^{-1}}$ gets cancelled, we get
\[\sin y=a\sin x\]
Now proceed with the differentiation.
In this also you will get the same answer, but we need to be very careful in this as the derivatives involve both $'x'$ and $'y'$ terms.