Question

If a trigonometric function is given as \[y=\sin \left( 2{{\sin }^{-1}}x \right)\], then show that \[\left( 1-{{x}^{2}} \right)\cdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=x\cdot \dfrac{dy}{dx}-4y\]

Answer 1

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 solution -
The given expression is,
\[y=\sin \left( 2{{\sin }^{-1}}x \right)........(i)\]
Multiplying both sides by ${{\sin }^{-1}}$ , we get
\[\Rightarrow {{\sin }^{-1}}y={{\sin }^{-1}}\sin (2{{\sin }^{-1}}x)\]
We know ${{\sin }^{-1}}\sin $ gets cancelled, so we get
\[\Rightarrow {{\sin }^{-1}}y=(2{{\sin }^{-1}}x)\]
Now differentiating with respect to $'x'$ , we get
\[\dfrac{d}{dx}({{\sin }^{-1}}y)=2\dfrac{d}{dx}({{\sin }^{-1}}x)\]
We know, \[\dfrac{d}{dx}({{\sin }^{-1}}x)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\], applying this formula the above equation can be written as,
\[\Rightarrow \dfrac{1}{\sqrt{1-{{y}^{2}}}}\cdot \dfrac{dy}{dx}=\dfrac{2}{\sqrt{1-{{x}^{2}}}}\]
By cross multiplying, we get
\[\Rightarrow \dfrac{dy}{dx}=2{{\left( \dfrac{1-{{y}^{2}}}{1-{{x}^{2}}} \right)}^{1/2}}\]
This is the first derivative, now by squaring both the sides, we get
\[\Rightarrow {{\left( \dfrac{dy}{dx} \right)}^{2}}=4\left( \dfrac{1-{{y}^{2}}}{1-{{x}^{2}}} \right)..........(ii)\]
Let us differentiate again to find the second order derivative, so differentiating the above expression with respect to $'x'$, we get
\[\Rightarrow \dfrac{d}{dx}\left( {{\left( \dfrac{dy}{dx} \right)}^{2}} \right)=4\dfrac{d}{dx}\left( \dfrac{1-{{y}^{2}}}{1-{{x}^{2}}} \right)\]
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
\[\Rightarrow 2\dfrac{dy}{dx}\cdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4\left( \dfrac{\left( 1-{{x}^{2}} \right)\dfrac{d}{dx}(1-{{y}^{2}})-(1-{{y}^{2}})\dfrac{d}{dx}(1-{{x}^{2}})}{{{\left( 1-{{x}^{2}} \right)}^{2}}} \right)\]
Applying the derivative, we get
\[\Rightarrow 2\dfrac{dy}{dx}\cdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=4\left( \dfrac{(1-{{x}^{2}})(-2y)\dfrac{dy}{dx}+(1-{{y}^{2}})2x}{{{\left( 1-{{x}^{2}} \right)}^{2}}} \right)\]
Multiplying both sides by $(1-{{x}^{2}})$ , we get
\[\begin{align}
  & \Rightarrow \left( 1-{{x}^{2}} \right)\cdot \dfrac{dy}{dx}\cdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\left( \dfrac{(1-{{x}^{2}})(-2y)\dfrac{dy}{dx}+2x(1-{{y}^{2}})}{1-{{x}^{2}}} \right) \\
 & \Rightarrow \left( 1-{{x}^{2}} \right)\cdot \dfrac{dy}{dx}\cdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-4y(1-{{x}^{2}})\dfrac{dy}{dx}+4x(1-{{y}^{2}})}{1-{{x}^{2}}} \\
\end{align}\]
Separating the terms, we get
\[\Rightarrow \left( 1-{{x}^{2}} \right)\cdot \dfrac{dy}{dx}\cdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-4y(1-{{x}^{2}})\dfrac{dy}{dx}}{1-{{x}^{2}}}+\dfrac{4x(1-{{y}^{2}})}{1-{{x}^{2}}}\]
Cancelling the like terms, we get
\[\Rightarrow \left( 1-{{x}^{2}} \right)\cdot \dfrac{dy}{dx}\cdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-4y\dfrac{dy}{dx}+x\left[ 4\dfrac{(1-{{y}^{2}})}{1-{{x}^{2}}} \right]\]
Now substituting value from equation (ii), w eget
\[\Rightarrow \left( 1-{{x}^{2}} \right)\cdot \dfrac{dy}{dx}\cdot \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-4y\dfrac{dy}{dx}+x{{\left( \dfrac{dy}{dx} \right)}^{2}}\]
Taking out the common term, we get
\[\Rightarrow \dfrac{dy}{dx}\left[ \left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right]=\dfrac{dy}{dx}\left[ -4y+x\left( \dfrac{dy}{dx} \right) \right]\]
Cancelling the like terms, we get
\[\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=x\dfrac{dy}{dx}-4y\]
Hence proved.

Note: In these types of questions first of all you have to check the given expression before starting the differentiation because sometimes the given expression can be in modified form.
Another way of solving this type of problem is finding the first order and second order derivatives, then substituting in \[\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=x\dfrac{dy}{dx}-4y\], separately and find out whether left hand side is equal to right hand side. In this method also you will get the same result.