Question

If $y = {\left( {x + \sqrt {\left( {1 + {x^2}} \right)} } \right)^n},$ then $\left( {1 + {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + x\dfrac{{dy}}{{dx}}{\text{ is;}}$
$\left( 1 \right){n^2}y$
$\left( 2 \right) - {n^2}y$
$\left( 3 \right) - y$
$\left( 4 \right)2{x^2}y$

Answer 1

This question deals with the application of standard derivative formulae. Some of them are listed here:
$\left( 1 \right)$ Power rule: $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ .
$\left( 2 \right)$ Derivative of a constant is always zero: $\dfrac{d}{{dx}}\left( a \right) = 0$ .
$\left( 3 \right)$ Product rule: $\dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}$ .
$\left( 4 \right)$ The quotient rule: $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}$ .
$\left( 5 \right)$ Chain rule of differentiation: This is used when we deal with composite functions i.e. functions within functions. For example: $\sin \left( {2{x^3} - 4x} \right)$ , in this example $\sin \left( x \right)$ is outer function and $\left( {2{x^3} - 4x} \right)$ is the inner function. Therefore, $\dfrac{d}{{dx}}\sin \left( {2{x^3} - 4x} \right) = \left( {6{x^2} - 4} \right)\cos \left( {2{x^3} - 4x} \right)$ .

Complete step by step solution:
The given expression for $y$ is ;
$y = {\left( {x + \sqrt {\left( {1 + {x^2}} \right)} } \right)^n}{\text{ }}......\left( 1 \right)$
$\left( {\because \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}} \right)$
Differentiating the above equation with respect to $x$ , we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\left( {x + \sqrt {\left( {1 + {x^2}} \right)} } \right)^n}$
Using the chain rule of differentiation, we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^{n - 1}} \times \left( {1 + \dfrac{1}{2}{{\left( {1 + {x^2}} \right)}^{\dfrac{{ - 1}}{2}}} \times 2x} \right)$
Simplifying the above equation, we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{n{{\left( {x + \sqrt {1 + {x^2}} } \right)}^n}}}{{x + \sqrt {1 + {x^2}} }} \times \left( {1 + \dfrac{{2x}}{{2\sqrt {1 + {x^2}} }}} \right)$
On further simplification;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{n{{\left( {x + \sqrt {1 + {x^2}} } \right)}^n}}}{{x + \sqrt {1 + {x^2}} }} \times \left( {\dfrac{{x + \sqrt {1 + {x^2}} }}{{\sqrt {1 + {x^2}} }}} \right)$
Cancelling the common term $x + \sqrt {1 + {x^2}}$ in numerator and denominator we get;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{n{{\left( {x + \sqrt {1 + {x^2}} } \right)}^n}}}{{\sqrt {1 + {x^2}} }}{\text{ }}......\left( 2 \right)$
On cross multiplication;
$\Rightarrow \dfrac{{dy}}{{dx}} \times \sqrt {1 + {x^2}} = n{\left( {x + \sqrt {1 + {x^2}} } \right)^n}$
Since $y = {\left( {x + \sqrt {\left( {1 + {x^2}} \right)} } \right)^n}$ , therefore replacing the value in the above equation we get;
$\Rightarrow \dfrac{{dy}}{{dx}} \times \sqrt {1 + {x^2}} = ny$
Squaring both the sides , we get ;
$\Rightarrow \left( {1 + {x^2}} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {n^2}{y^2}{\text{ }}......\left( 3 \right)$
Differentiating the above equation, we get;
By the product rule of differentiation, we know that;
$\left[ {\because \dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}} \right]$
Let $u = \left( {1 + {x^2}} \right){\text{ and }}v = {\left( {\dfrac{{dy}}{{dx}}} \right)^2}$
$\Rightarrow \left( {2x} \right){\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \left( {1 + {x^2}} \right) \times 2\dfrac{{dy}}{{dx}} \times {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = {n^2} \times 2y\dfrac{{dy}}{{dx}}$
Further simplifying the above equation , we get;
$\Rightarrow 2x \times \dfrac{{{d^2}y}}{{d{x^2}}} + \left( {1 + {x^2}} \right)2\left( {\dfrac{{dy}}{{dx}}} \right)\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right) = 2{n^2}y\dfrac{{dy}}{{dx}}$
As we have to calculate the value of $\left( {1 + {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + x\dfrac{{dy}}{{dx}}$ according to the given question , therefore, rearranging the terms to get the desired expression ;
Taking the term $2\dfrac{{dy}}{{dx}}$ outside on the L.H.S. , we get ;
$\Rightarrow 2\left( {\dfrac{{dy}}{{dx}}} \right)\left( {\left( {1 + {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + x\dfrac{{dy}}{{dx}}} \right) = 2{n^2}y\dfrac{{dy}}{{dx}}$
Further simplifying the above expression;
$\Rightarrow \left( {1 + {x^2}} \right)\dfrac{{{d^2}y}}{{d{x^2}}} + x\dfrac{{dy}}{{dx}} = {n^2}y$
Therefore, the correct answer for this question is option $\left( 1 \right)$ i.e. ${n^2}y$.

Note:
We should always simplify the given expression first, before differentiation and always try to proceed according to the expression that has been given to us. There are chances of making a mistake while differentiating, whenever a square root is given in the expression but we know that $\sqrt x = {\left( x \right)^{\dfrac{1}{2}}}$ so, always keep this in mind. With the basic knowledge of algebraic identities and standard derivative formulae of functions , this type of questions can be solved very easily.