Question

If $y=\sin \left( m{{\sin }^{-1}}x \right)$, then $\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+{{m}^{2}}y$ is ?

Answer 1

In this problem, we have to find the value of the given derivative expression $\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+{{m}^{2}}y$ if $y=\sin \left( m{{\sin }^{-1}}x \right)$. We can first take the given condition $y=\sin \left( m{{\sin }^{-1}}x \right)$, we can differentiate it on both sides, we can again find the second derivative using the $\dfrac{u}{v}$ method, where $\dfrac{u}{v}=\dfrac{u'v-uv'}{{{v}^{2}}}$. We can then simplify the steps by substituting the assumed values and find the value.

Complete step by step answer:
Here we have to find the value of $\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+{{m}^{2}}y$.
We are given that,
$\Rightarrow y=\sin \left( m{{\sin }^{-1}}x \right)$…….. (1)
We can now differentiate on both sides, we get
$\Rightarrow \dfrac{dy}{dx}=\cos \left( m{{\sin }^{-1}}x \right)\times \dfrac{m}{\sqrt{1-{{x}^{2}}}}$
We can now simplify the above step, we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{m\cos \left( m{{\sin }^{-1}}x \right)}{\sqrt{1-{{x}^{2}}}}$……….. (2)
We can now find the second derivative using the $\dfrac{u}{v}$ method, where $\dfrac{u}{v}=\dfrac{u'v-uv'}{{{v}^{2}}}$.
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=m\left[ \dfrac{\left( \sqrt{1-{{x}^{2}}} \right)\left( -\sin \left( m\left( {{\sin }^{-1}}x \right) \right) \right)\times \dfrac{m}{\sqrt{1-{{x}^{2}}}}-\left( \cos \left( m\left( {{\sin }^{-1}}x \right) \right) \right)\times \dfrac{1}{2\sqrt{1-{{x}^{2}}}}\times \left( 0-2x \right)}{{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}} \right]$
We can now simplify the above step by cancelling similar terms, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=m\left[ \dfrac{\left( -m\sin \left( m\left( {{\sin }^{-1}}x \right) \right) \right)+\left( \cos \left( m\left( {{\sin }^{-1}}x \right) \right) \right)\times \dfrac{x}{\sqrt{1-{{x}^{2}}}}}{1-{{x}^{2}}} \right]$
We can now write the denominator separately for each term, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=m\left[ \dfrac{\left( -m\sin \left( m\left( {{\sin }^{-1}}x \right) \right) \right)}{1-{{x}^{2}}}+\dfrac{\left( x\cos \left( m\left( {{\sin }^{-1}}x \right) \right) \right)}{1-{{x}^{2}}\sqrt{1-{{x}^{2}}}} \right]$
We can now substitute the derivative (2), we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=m\left[ \dfrac{\left( -m\sin \left( m\left( {{\sin }^{-1}}x \right) \right) \right)}{1-{{x}^{2}}}+\dfrac{x\dfrac{dy}{dx}}{1-{{x}^{2}}} \right]$
We can see that we have common denominator now, so we can multiply $1-{{x}^{2}}$ on both sides, we get
$\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=m\left[ \left( -m\sin \left( m\left( {{\sin }^{-1}}x \right) \right) \right)+x\dfrac{dy}{dx} \right]$
We can now multiply the term m insides, we get
$\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left[ \left( -{{m}^{2}}\sin \left( m\left( {{\sin }^{-1}}x \right) \right) \right)+mx\dfrac{dy}{dx} \right]$
We can now substitute (1), in the above step, we get
$\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-{{m}^{2}}y+mx\dfrac{dy}{dx}$
We can now add $-{{m}^{2}}y+mx\dfrac{dy}{dx}$ on both sides, we get
$\Rightarrow \left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-mx\dfrac{dy}{dx}+{{m}^{2}}y=0$
Therefore, the value of $\left( 1-{{x}^{2}} \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-x\dfrac{dy}{dx}+{{m}^{2}}y$ is 0.

Note: We should always remember that, if we have terms in fraction, we can differentiate it using the $\dfrac{u}{v}$ method, where $\dfrac{u}{v}=\dfrac{u'v-uv'}{{{v}^{2}}}$. We should also know the differentiation formula to differentiate the terms and simplify to find the required answer.