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Question: Show that \[y = {\left( {{{\sin }^{ - 1}}x} \right)^2}\] then \[\left( {1 - {x^2}} \right)\dfrac{{{d...

Show that y=(sin⁡−1x)2y = {\left( {{{\sin }^{ - 1}}x} \right)^2} then (1−x2)d2ydx2−xdydx−2=0\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} - 2 = 0?

Explanation

Solution

In the question we have to first derive the condition, and now again we have to differentiate the result, i.e., we will get the second derivative, and now substituting the first and second derivative values in the expression we will get the required result.

Complete step-by-step answer:
Given y=(sin⁡−1x)2y = {\left( {{{\sin }^{ - 1}}x} \right)^2},
Now differentiating both sides with respect of xx,
⇒dydx=ddx(sin⁡−1x)2\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\left( {{{\sin }^{ - 1}}x} \right)^2},
Now using the derivative identity ddxx2=2x\dfrac{d}{{dx}}{x^2} = 2x and using the chain rule we get, we get,
⇒dydx=2sin⁡−1xddxsin⁡−1x\Rightarrow \dfrac{{dy}}{{dx}} = 2{\sin ^{ - 1}}x\dfrac{d}{{dx}}{\sin ^{ - 1}}x,
Now again derivativation we get,
⇒dydx=2sin⁡−1x(11−x2)\Rightarrow \dfrac{{dy}}{{dx}} = 2{\sin ^{ - 1}}x\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right),
Now again differentiating we get,
⇒d2ydx2=ddx(2sin⁡−1x(11−x2))\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {2{{\sin }^{ - 1}}x\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right)} \right),
Now using quotient rule, f′(x)=u′(x)×v(x)−u(x)v′(x)(v(x))2f'\left( x \right) = \dfrac{{u'\left( x \right) \times v\left( x \right) - u\left( x \right)v'\left( x \right)}}{{{{\left( {v\left( x \right)} \right)}^2}}}, we get,
⇒d2ydx2=2(11−x2(ddxsin⁡−1x)−sin⁡−1x(ddx11−x2)(1−x2)2)\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 2\left( {\dfrac{{\dfrac{1}{{\sqrt {1 - {x^2}} }}\left( {\dfrac{d}{{dx}}{{\sin }^{ - 1}}x} \right) - {{\sin }^{ - 1}}x\left( {\dfrac{d}{{dx}}\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right)}}{{{{\left( {\sqrt {1 - {x^2}} } \right)}^2}}}} \right),
Now deriving we get,
⇒d2ydx2=(2(1−x2)(11−x2)−sin⁡−1x(12(−2x)1−x2)(1−x2))\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \left( {\dfrac{{2\left( {\sqrt {1 - {x^2}} } \right)\left( {\dfrac{1}{{\sqrt {1 - {x^2}} }}} \right) - {{\sin }^{ - 1}}x\left( {\dfrac{{\dfrac{1}{2}\left( { - 2x} \right)}}{{\sqrt {1 - {x^2}} }}} \right)}}{{\left( {1 - {x^2}} \right)}}} \right),
Now simplifying we get,
⇒d2ydx2=(2−sin⁡−1x((−x)1−x2)(1−x2))\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \left( {\dfrac{{2 - {{\sin }^{ - 1}}x\left( {\dfrac{{\left( { - x} \right)}}{{\sqrt {1 - {x^2}} }}} \right)}}{{\left( {1 - {x^2}} \right)}}} \right),
Now taking out the minus sign we get,
⇒d2ydx2=(2+sin⁡−1x(x1−x2)(1−x2))\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \left( {\dfrac{{2 + {{\sin }^{ - 1}}x\left( {\dfrac{x}{{\sqrt {1 - {x^2}} }}} \right)}}{{\left( {1 - {x^2}} \right)}}} \right),
Now taking (1−x2)\left( {1 - {x^2}} \right) to the Left hand side we get,
⇒(1−x2)d2ydx2=(2+sin⁡−1x(x1−x2))\Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = \left( {2 + {{\sin }^{ - 1}}x\left( {\dfrac{x}{{\sqrt {1 - {x^2}} }}} \right)} \right),
Now substituting dydx\dfrac{{dy}}{{dx}} value in the above equation we get,
⇒(1−x2)d2ydx2=(2+xdydx)\Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = \left( {2 + x\dfrac{{dy}}{{dx}}} \right),
Now simplifying we get,
⇒(1−x2)d2ydx2=2+xdydx\Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = 2 + x\dfrac{{dy}}{{dx}},
Now taking terms in right hand side to left hand side we get,
⇒(1−x2)d2ydx2−xdydx−2=0\Rightarrow \left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} - 2 = 0,
Hence proved.
Also considering the left hand side of the given expression and substituting the values in the expression we get,
⇒(1−x2)(2+2xsin⁡−1x1−x21−x2)−x(2sin⁡−1x1−x2)−2\Rightarrow \left( {1 - {x^2}} \right)\left( {\dfrac{{2 + \dfrac{{2x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}}}{{1 - {x^2}}}} \right) - x\left( {\dfrac{{2{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}} \right) - 2,
Now simplifying we get,
⇒2+2xsin⁡−1x1−x2−2xsin⁡−1x1−x2−2\Rightarrow 2 + \dfrac{{2x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }} - \dfrac{{2x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }} - 2,
Now eliminating the like terms we get,
⇒2+2xsin⁡−1x1−x2−2xsin⁡−1x1−x2−2=0\Rightarrow 2 + \dfrac{{2x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }} - \dfrac{{2x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }} - 2 = 0.
Final Answer:

∴\therefore Ify=(sin⁡−1x)2y = {\left( {{{\sin }^{ - 1}}x} \right)^2} then (1−x2)d2ydx2−xdydx−2=0\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - x\dfrac{{dy}}{{dx}} - 2 = 0. Hence proved.

Note:
Differentiation is the method of evaluating a function’s derivative at any time. Some of the fundamental rules for differentiation are given below, and using these rules we can solve differentiation questions easily.
Sum or difference rule: When the function is the sum or difference of two functions, the derivative is the sum or difference of derivative of each function, i.e.,
If f(x)=u(x)±v(x)f\left( x \right) = u\left( x \right) \pm v\left( x \right),
Then f′(x)=u′(x)±v′(x)f'\left( x \right) = u'\left( x \right) \pm v'\left( x \right).
Product rule: When the function is the product of two functions, the derivative is the sum or difference of derivative of each function, i.e.,
If f(x)=u(x)×v(x)f\left( x \right) = u\left( x \right) \times v\left( x \right),
Then f′(x)=u′(x)v(x)+u(x)v′(x)f'\left( x \right) = u'\left( x \right)v\left( x \right) + u\left( x \right)v'\left( x \right)
Quotient Rule: If the function is in the form of two functions u(x)v(x)\dfrac{{u\left( x \right)}}{{v\left( x \right)}}, the derivative of the function can be expressed as,
f′(x)=u′(x)v(x)−u(x)v′(x)(v(x))2f'\left( x \right) = \dfrac{{u'\left( x \right)v\left( x \right) - u\left( x \right)v'\left( x \right)}}{{{{\left( {v\left( x \right)} \right)}^2}}}.
Chain Rule:
If y=f(x)=g(u)y = f\left( x \right) = g\left( u \right),
And if u=h(x)u = h\left( x \right),
Then dydx=dydu×dudx\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}