Question

If $y=\text{cose}{{\text{c}}^{-1}}x,x>1$, then show that $x\left( {{x}^{2}}-1 \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\left( 2{{x}^{2}}-1 \right)\dfrac{dy}{dx}=0$

Answer 1

Hint: We are given a function as: $y=\text{cose}{{\text{c}}^{-1}}x,x>1$. So, differentiation the given function with respect to x to get the value of $\dfrac{dy}{dx}$ by using the formula: $\dfrac{d}{dx}\left( \text{cose}{{\text{c}}^{-1}}x \right)=\dfrac{-1}{x\sqrt{{{x}^{2}}-1}}$.
Then, differentiate $\dfrac{dy}{dx}$ with respect to x and get the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ by using quotient rule of differentiation, i.e. $\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\dfrac{f'(x)g(x)-f(x)g'(x)}{{{g}^{2}}(x)}$. Now, substitute the values of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ and $\dfrac{dy}{dx}$ in the equation: $x\left( {{x}^{2}}-1 \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\left( 2{{x}^{2}}-1 \right)\dfrac{dy}{dx}$ and prove that the equation is equals to zero.

Complete step by step answer:
Since we have a function: $y=\text{cose}{{\text{c}}^{-1}}x......(1)$
Now, using the formula $\dfrac{d}{dx}\left( \text{cose}{{\text{c}}^{-1}}x \right)=\dfrac{-1}{x\sqrt{{{x}^{2}}-1}}$, differentiate equation (1) with respect to x, we get:
$\dfrac{dy}{dx}=\dfrac{-1}{x\sqrt{{{x}^{2}}-1}}......(2)$
So, we have the value of $\dfrac{dy}{dx}$.
Now, to get the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ , differentiate equation (2) with respect to x, we get:
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{-1}{x\sqrt{{{x}^{2}}-1}} \right)......(3)$
As we know, by quotient rule of differentiation:
$\dfrac{d}{dx}\left( \dfrac{f(x)}{g(x)} \right)=\dfrac{f'(x)g(x)-f(x)g'(x)}{{{g}^{2}}(x)}$
By using this formula, we can write equation (3) as:
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{\left( \dfrac{d}{dx}(-1)\times x\sqrt{{{x}^{2}}-1} \right)-\left( (-1)\times \dfrac{d}{dx}\left( x\sqrt{{{x}^{2}}-1} \right) \right)}{{{\left( x\sqrt{{{x}^{2}}-1} \right)}^{2}}} \right)......(3)$
Since we know that:
$\dfrac{d}{dx}\left( f(x)g(x) \right)=f'(x)g(x)+f(x)g(x)$
So, we can write equation (3) as:
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{\left( \dfrac{d}{dx}(-1)\times x\sqrt{{{x}^{2}}-1} \right)-\left( (-1)\times \left( \left( \dfrac{d}{dx}x \right)\sqrt{{{x}^{2}}-1}+x\left( \dfrac{d}{dx}\sqrt{{{x}^{2}}-1} \right) \right) \right)}{{{\left( x\sqrt{{{x}^{2}}-1} \right)}^{2}}} \right)......(4)\]
As we know that:
$\left[ \dfrac{d}{dx}a=0;\dfrac{d}{dx}x=1;\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}};\dfrac{d}{dx}f(g(x))=f'(g(x))g'(x) \right]$
By applying the above differentiation rules in equation (4), we get:
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{\left( 0\times x\sqrt{{{x}^{2}}-1} \right)-\left( (-1)\times \left( 1\times \sqrt{{{x}^{2}}-1}+x\left( \dfrac{1}{2\sqrt{{{x}^{2}}-1}}\times 2x \right) \right) \right)}{{{\left( x\sqrt{{{x}^{2}}-1} \right)}^{2}}} \right) \\
 & =\left( \dfrac{-\left( (-1)\times \left( \sqrt{{{x}^{2}}-1}+\dfrac{{{x}^{2}}}{\sqrt{{{x}^{2}}-1}} \right) \right)}{{{x}^{2}}\left( {{x}^{2}}-1 \right)} \right)
\end{align}\]
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{\sqrt{{{x}^{2}}-1}+\dfrac{{{x}^{2}}}{\sqrt{{{x}^{2}}-1}}}{{{x}^{2}}\left( {{x}^{2}}-1 \right)} \right) \\
 & =\dfrac{{{\left( \sqrt{{{x}^{2}}-1} \right)}^{2}}+{{x}^{2}}}{{{x}^{2}}\left( {{x}^{2}}-1 \right)\sqrt{{{x}^{2}}-1}}
\end{align}\]
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{x}^{2}}-1+{{x}^{2}}}{{{x}^{2}}\left( {{x}^{2}}-1 \right)\sqrt{{{x}^{2}}-1}} \\
 & =\dfrac{2{{x}^{2}}-1}{{{x}^{2}}\left( {{x}^{2}}-1 \right)\sqrt{{{x}^{2}}-1}}......(5)
\end{align}\]
Now, we have value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ and $\dfrac{dy}{dx}$.
Substitute the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ and $\dfrac{dy}{dx}$ from equation (2) and equation (5) in the equation:
$x\left( {{x}^{2}}-1 \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\left( 2{{x}^{2}}-1 \right)\dfrac{dy}{dx}$, we get:
$\begin{align}
  & \Rightarrow x\left( {{x}^{2}}-1 \right)\left( \dfrac{2{{x}^{2}}-1}{{{x}^{2}}\left( {{x}^{2}}-1 \right)\sqrt{{{x}^{2}}-1}} \right)+\left( 2{{x}^{2}}-1 \right)\left( \dfrac{-1}{x\sqrt{{{x}^{2}}-1}} \right) \\
 & \Rightarrow \dfrac{2{{x}^{2}}-1}{x\sqrt{{{x}^{2}}-1}}-\dfrac{2{{x}^{2}}-1}{x\sqrt{{{x}^{2}}-1}} \\
 & \Rightarrow 0=RHS \\
\end{align}$

Hence, proved that $x\left( {{x}^{2}}-1 \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\left( 2{{x}^{2}}-1 \right)\dfrac{dy}{dx}=0$

Note: Always go step-by-step to solve a complex differentiation problem because there are many rules applied in the above solution. So, you might get confused while applying the rules of differentiation. This will make a simple solution messed up. So, choose the correct rule to apply for solving a derivative and do not apply all the rules at one time. Avoid mixing up formulas and always try to solve the problem with an easier rule.