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

Hint: 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.