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Question: If \[y=\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\], prove that: \[(1-{{x}^{2}})\dfrac{dy}{dx}=x+\...

If y=xsin⁡−1x1−x2y=\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}, prove that: (1−x2)dydx=x+yx(1-{{x}^{2}})\dfrac{dy}{dx}=x+\dfrac{y}{x}

Explanation

Solution

Hint: To prove this we can differentiate the function w.r.t x by using product rule and quotient rule.
ddx(u.v)=u.ddx(v)+v.ddx(u)\dfrac{d}{dx}(u.v)=u.\dfrac{d}{dx}(v)+v.\dfrac{d}{dx}(u) [Product Rule]
ddx(uv)=v.dudx−u.dvdxv2\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v.\dfrac{du}{dx}-u.\dfrac{dv}{dx}}{\mathop{v}^{2}} [Quotient Rule]
Complete step-by-step answer:
Given y=xsin⁡−1x1−x2y=\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}
Differentiating the function w.r.t. x
dydx=ddx[xsin⁡−1x1−x2]\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}} \right]
Using (uv)(uv)rule and (uv)\left( \dfrac{u}{v} \right)rule,
dydx=1−x2.ddx(xsin⁡−1x)−(xsin⁡−1x)ddx1−x2(1−x2)2\dfrac{dy}{dx}=\dfrac{\sqrt{1-{{x}^{2}}}.\dfrac{d}{dx}(x{{\sin }^{-1}}x)-(x{{\sin }^{-1}}x)\dfrac{d}{dx}\sqrt{1-{{x}^{2}}}}{\mathop{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}
Now, ddx(xsin⁡−1x)=x.11−x2+sin⁡−1x(1)\dfrac{d}{dx}(x{{\sin }^{-1}}x)=x.\dfrac{1}{\sqrt{1-{{x}^{2}}}}+{{\sin }^{-1}}x(1)
\left\\{ \dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right\\}
Now,
ddx1−x2=121−x2×(−2x)⇒−x1−x2\dfrac{d}{dx}\sqrt{1-{{x}^{2}}}=\dfrac{1}{2\sqrt{1-{{x}^{2}}}}\times( -2x)\Rightarrow \dfrac{-x}{\sqrt{1-{{x}^{2}}}}
\left\\{ \dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}} \right\\}
∴dydx=1−x2.[x1−x2+sin⁡−1x]−xsin⁡−1x−x1−x2(1−x2)\therefore \dfrac{dy}{dx}=\dfrac{\sqrt{1-{{x}^{2}}}.\left[ \dfrac{x}{\sqrt{1-{{x}^{2}}}}+{{\sin }^{-1}}x \right]-x{{\sin }^{-1}}x\dfrac{-x}{\sqrt{1-{{x}^{2}}}}}{(1-{{x}^{2}})}
∴dydx=x+sin⁡−1x.(1−x2)−x2sin⁡−1x1−x2(1−x2)\therefore \dfrac{dy}{dx}=\dfrac{x+{{\sin }^{-1}}x.(\sqrt{1-{{x}^{2}}})-\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}{(1-{{x}^{2}})}
Taking (1−x2)(1-{{x}^{2}})to the other side,
⇒(1−x2)dydx=x+sin⁡−1x.(1−x2)−x2sin⁡−1x1−x2\Rightarrow (1-{{x}^{2}})\dfrac{dy}{dx}=x+{{\sin }^{-1}}x.(\sqrt{1-{{x}^{2}}})-\dfrac{{{x}^{2}}{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}
⇒(1−x2)dydx=x+(1−x2)2sin⁡−1x+x2sin⁡−1x1−x2\Rightarrow (1-{{x}^{2}})\dfrac{dy}{dx}=x+\dfrac{\mathop{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}{{\sin }^{-1}}x+{{x}^{2}}{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}
⇒(1−x2)dydx=x+(1−x2)sin⁡−1x+x2sin⁡−1x1−x2\Rightarrow (1-\mathop{x}^{2})\dfrac{dy}{dx}=x+\dfrac{(1-\mathop{x}^{2}){{\sin }^{-1}}x+{{x}^{2}}{{\sin }^{-1}}x}{\sqrt{1-\mathop{x}^{2}}}
⇒(1−x2)dydx=x+sin⁡−1x−x2sin⁡−1x+x2sin⁡−1x1−x2\Rightarrow (1-\mathop{x}^{2})\dfrac{dy}{dx}=x+\dfrac{{{\sin }^{-1}}x-{{x}^{2}}{{\sin }^{-1}}x+{{x}^{2}}{{\sin }^{-1}}x}{\sqrt{1-\mathop{x}^{2}}}
⇒(1−x2)dydx=x+sin⁡−1x1−x2\Rightarrow (1-\mathop{x}^{2})\dfrac{dy}{dx}=x+\dfrac{{{\sin }^{-1}}x}{\sqrt{1-\mathop{x}^{2}}}
Given y=xsin⁡−1x1−x2⇒sin⁡−1x1−x2=yxy=\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-\mathop{x}^{2}}}\Rightarrow \dfrac{{{\sin }^{-1}}x}{\sqrt{1-\mathop{x}^{2}}}=\dfrac{y}{x}
∴\therefore Substituting the value of sin⁡−1x1−x2\dfrac{{{\sin }^{-1}}x}{\sqrt{1-\mathop{x}^{2}}} as yx\dfrac{y}{x} in eq. 1
(1−x2)dydx=x+yx(1-\mathop{x}^{2})\dfrac{dy}{dx}=x+\dfrac{y}{x}
Hence proved.
Note: In this question we need to be careful about the order of application of formula. First we need to use the quotient rule and then we need to use the multiplication rule.