Question

If we have an expression as $y={{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{15}}+{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{15}}$ then $\left( {{x}^{2}}-1 \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}$ is equal to
(a) 125y
(b) $225{{y}^{2}}$
(c) 225y
(d) $224{{y}^{2}}$

Answer 1

First, before proceeding for this, we must know the rule of the differentiation which is a chain rule that states that differentiation of x is continued till we get the single term of x. Then, by applying the above-stated rule to the given function and we calculate the first derivative. Then, by again differentiating the above equation with respect to x by using the product rule in differentiation, we get the desired result after arrangement and substitution.

Complete step-by-step solution:
In this question, we are supposed to find the value of $\left( {{x}^{2}}-1 \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}$when it is given that $y={{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{15}}+{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{15}}$.
So, before proceeding for this, we must know the rule of the differentiation which is a chain rule that states that differentiation of x is continued till we get the single term of x.
Now, by applying the above stated rule to the given function and we calculate the first derivative as:
$\begin{aligned} & \dfrac{dy}{dx}=15{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{14}}\times \dfrac{d}{dx}\left( x+\sqrt{{{x}^{2}}-1} \right)+15{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{14}}\times \dfrac{d}{dx}\left( x-\sqrt{{{x}^{2}}-1} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=15{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{14}}\times \left( 1+\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)+15{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{14}}\times \left( 1-\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right) \\\ \end{aligned}$
Now, by solving the first derivative of function by taking LCM, we get:
$\begin{aligned} & \dfrac{dy}{dx}=15{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{14}}\times \left( \dfrac{\sqrt{{{x}^{2}}-1}+x}{\sqrt{{{x}^{2}}-1}} \right)+15{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{14}}\times \left( \dfrac{\sqrt{{{x}^{2}}-1}-x}{\sqrt{{{x}^{2}}-1}} \right) \\\ & \Rightarrow \dfrac{dy}{dx}=15{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{15}}\times \left( \dfrac{1}{\sqrt{{{x}^{2}}-1}} \right)-15{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{15}}\times \left( \dfrac{1}{\sqrt{{{x}^{2}}-1}} \right) \\\ \end{aligned}$
Now, by rearranging the above expression, we get:
$\sqrt{{{x}^{2}}-1}\times \dfrac{dy}{dx}=15{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{15}}-15{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{15}}$
Then, by again differentiating the above equation with respect to x by using the product rule in differentiation, we get:
$\sqrt{{{x}^{2}}-1}\times \dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\dfrac{dy}{dx}\times \dfrac{x}{\sqrt{{{x}^{2}}-1}}=15\times 15\dfrac{{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{15}}}{\sqrt{{{x}^{2}}-1}}+15\times 15\dfrac{{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{15}}}{\sqrt{{{x}^{2}}-1}}$
Now, by taking the LCM as $\sqrt{{{x}^{2}}-1}$on both sides and solving, we get:
$\left( {{x}^{2}}-1 \right)\times \dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}=225{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{15}}+225{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{15}}$
Then, by taking 225 common from the right hand side of the expression and by substituting the value as given in question as $y={{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{15}}+{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{15}}$, we get:
$\left( {{x}^{2}}-1 \right)\times \dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}=225y$
So, we get the value of the expression $\left( {{x}^{2}}-1 \right)\times \dfrac{{{d}^{2}}y}{d{{x}^{2}}}+x\dfrac{dy}{dx}$as 225y.
Hence, the option (c) is correct.

Note: Now, to solve these types of the questions we need to know some of the basic rules of differentiation to get the answer. So some of the basic rules are as follows:
Product rule in differentiation is as $\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$ where u and v are different functions.
Also one basic differentiation used is $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ where n can be any number.