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Question: If \(y=\text{cose}{{\text{c}}^{-1}}x,x>1\), then show that \(x\left( {{x}^{2}}-1 \right)\dfrac{{{d}^...

If y=cosec−1x,x>1y=\text{cose}{{\text{c}}^{-1}}x,x>1, then show that x(x2−1)d2ydx2+(2x2−1)dydx=0x\left( {{x}^{2}}-1 \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\left( 2{{x}^{2}}-1 \right)\dfrac{dy}{dx}=0

Explanation

Solution

We are given a function as: y=cosec−1x,x>1y=\text{cose}{{\text{c}}^{-1}}x,x>1. So, differentiation the given function with respect to x to get the value of dydx\dfrac{dy}{dx} by using the formula: ddx(cosec−1x)=−1xx2−1\dfrac{d}{dx}\left( \text{cose}{{\text{c}}^{-1}}x \right)=\dfrac{-1}{x\sqrt{{{x}^{2}}-1}}.
Then, differentiate dydx\dfrac{dy}{dx} with respect to x and get the value of d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}} by using quotient rule of differentiation, i.e. ddx(f(x)g(x))=f′(x)g(x)−f(x)g′(x)g2(x)\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 d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}} and dydx\dfrac{dy}{dx} in the equation: x(x2−1)d2ydx2+(2x2−1)dydxx\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=cosec−1x......(1)y=\text{cose}{{\text{c}}^{-1}}x......(1)
Now, using the formula ddx(cosec−1x)=−1xx2−1\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:
dydx=−1xx2−1......(2)\dfrac{dy}{dx}=\dfrac{-1}{x\sqrt{{{x}^{2}}-1}}......(2)
So, we have the value of dydx\dfrac{dy}{dx}.
Now, to get the value of d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}} , differentiate equation (2) with respect to x, we get:
d2ydx2=ddx(−1xx2−1)......(3)\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:
ddx(f(x)g(x))=f′(x)g(x)−f(x)g′(x)g2(x)\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:
d2ydx2=((ddx(−1)×xx2−1)−((−1)×ddx(xx2−1))(xx2−1)2)......(3)\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:
ddx(f(x)g(x))=f′(x)g(x)+f(x)g(x)\dfrac{d}{dx}\left( f(x)g(x) \right)=f'(x)g(x)+f(x)g(x)
So, we can write equation (3) as:
d2ydx2=((ddx(−1)×xx2−1)−((−1)×((ddxx)x2−1+x(ddxx2−1)))(xx2−1)2)......(4)\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:
[ddxa=0;ddxx=1;ddxx=12x;ddxf(g(x))=f′(g(x))g′(x)]\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:

& \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{aligned}$$ $$\begin{aligned} & \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{aligned}$$ $$\begin{aligned} & \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{aligned}$$ 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: $\begin{aligned} & \Rightarrow x\left( {{x}^{2}}-1 \right)\left( \dfrac{2{{x}^{2}}-1}{{{x}^{2}}\left( {{x}^{2}}-1 \right)\sqrt{{{x}^{2}}-1}} \right)+\left( 2{{x}^{2}}-1 \right)\left( \dfrac{-1}{x\sqrt{{{x}^{2}}-1}} \right) \\\ & \Rightarrow \dfrac{2{{x}^{2}}-1}{x\sqrt{{{x}^{2}}-1}}-\dfrac{2{{x}^{2}}-1}{x\sqrt{{{x}^{2}}-1}} \\\ & \Rightarrow 0=RHS \\\ \end{aligned}$ **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.