Question

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

Answer 1

Hint : Find the double derivative of the given trigonometric equation and after reducing the equation find its value.
In this question we are given with a trigonometric equation and since the equation whose value is to be found is in second derivative so we will differentiate the given equation two times with respect to x to find the value of the equation.
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}\left( y \right)} \right)$

Complete step-by-step answer :
Given the trigonometric equation $y = \sin \left( {m{{\sin }^{ - 1}}x} \right) - - (i)$
Now we differentiate the given equation (i) with respect to $x$ by using the chain rule, hence we get

$$\dfrac{{d\left( y \right)}}{{dx}} = \dfrac{{d\left( {\sin \left( {m{{\sin }^{ - 1}}x} \right)} \right)}}{{dx}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \cos \left( {m\left( {{{\sin }^{ - 1}}x} \right)} \right) \times \dfrac{m}{{\sqrt {1 - {x^2}} }} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{m\cos \left( {m\left( {{{\sin }^{ - 1}}x} \right)} \right)}}{{\sqrt {1 - {x^2}} }} - - (ii) \\\$$

Now since the equation whose value is to be found is in second derivative form so we need to again differentiate the equation to bring it in double derivative form, so we differentiate the equation with respect to $x$ by using quotient rule of derivation, we can write
$\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{m\cos \left( {m\left( {{{\sin }^{ - 1}}x} \right)} \right)}}{{\sqrt {1 - {x^2}} }}} \right)$
Hence by differentiating the above equation, we get

$$\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] \\\ \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] \\\ \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\left( {\cos \left( {m\left( {{{\sin }^{ - 1}}x} \right)} \right)} \right)}}{{\left( {1 - {x^2}} \right)\sqrt {1 - {x^2}} }}} \right] \\\$$

Now as we already know $\dfrac{{dy}}{{dx}} = \dfrac{{m\cos \left( {m\left( {{{\sin }^{ - 1}}x} \right)} \right)}}{{\sqrt {1 - {x^2}} }}$ from equation (ii), so we can further write this equation as
$\dfrac{{{d^2}y}}{{d{x^2}}} = m\left[ {\dfrac{{\left( { - m\sin \left( {m\left( {{{\sin }^{ - 1}}x} \right)} \right)} \right)}}{{1 - {x^2}}} + \dfrac{{x\dfrac{{dy}}{{dx}}}}{{\left( {1 - {x^2}} \right)}}} \right]$
Now we cross multiply the obtained equation, we can write
$\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = mx\dfrac{{dy}}{{dx}} - {m^2}\sin \left( {m\left( {{{\sin }^{ - 1}}x} \right)} \right)$
Now as $y = \sin \left( {m{{\sin }^{ - 1}}x} \right)$ given in the question, we can further write this equation as
$\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} = mx\dfrac{{dy}}{{dx}} - {m^2}y$
Now this equation can also be written in the form of
$\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - mx\dfrac{{dy}}{{dx}} + {m^2}y = 0$
Therefore we can say the values of the derivative $\left( {1 - {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} - mx\dfrac{{dy}}{{dx}} + {m^2}y = 0$
Hence the value of the derivative is 0.
So, the correct answer is “0”.

Note : Important method of derivation used:
Chain Rule: $\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$
Quotient rule: $\dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left[ {g\left( x \right)} \right] }^2}}}$
Double derivatives are basically used to determine the nature of the stationary objects
whether the point is maximum, minimum or points of inflation. In this question since the equation whose value was to be found was in second order so we needed the double derivative of the equation usually represented as:
$\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}\left( y \right)} \right)$