Question

Find \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]
\[y=\dfrac{1+x}{1-x}\]

Answer 1

Hint: If u and v are two differentiable functions of x then \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\times \dfrac{du}{dx}-u\times \dfrac{dv}{dx}}{{{v}^{2}}}\]and this formula is called quotient rule. In this problem u and v are two differentiable functions of x so apply the quotient rule. Here they asked us to find the \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]so we have to apply the quotient rule two times then we will get the required answer.

Complete step-by-step answer:
Given that \[y=\dfrac{1+x}{1-x}\]
We have to find \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]
We know that the formula of \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)\]is given by \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\times \dfrac{du}{dx}-u\times \dfrac{dv}{dx}}{{{v}^{2}}}\]
Applying the above formula we will get,
\[\dfrac{dy}{dx}=\dfrac{(1-x)\left( 1 \right)-\left( 1+x \right)\left( -1 \right)}{{{(1-x)}^{2}}}\]. . . . . . . . . . . . . . . . . . .(1)
\[\dfrac{dy}{dx}=\dfrac{1-x+1+x}{{{(1-x)}^{2}}}\]
\[\dfrac{dy}{dx}=\dfrac{2}{{{(1-x)}^{2}}}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(2)
Now again apply the derivative then we will get the double derivative of it.
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)\]
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{2}{{{(1-x)}^{2}}} \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . (3)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{(1-x)}^{2}}\times 0-2\times 2\times (1-x)\left( -1 \right)}{{{(1-x)}^{4}}}\]. . . . . . . . . . . . (4)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2\times 2\times (1-x)}{{{(1-x)}^{4}}}\] . . . . . . . . . . . . . . . . . . . . . . . . . . . (5)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{4}{{{(1-x)}^{3}}}\]

Note:The formula for \[\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}\]. So the derivative of \[1+x\]is 1and this formula is known as sum rule. The formula for derivative of \[{{u}^{n}}=n{{u}^{n-1}}\]where n is any integer so the derivative of \[{{(1-x)}^{2}}\]is \[2\times (1-x)\left( -1 \right)\]. Carefully do the basic mathematical operations like addition subtraction then we will get the required answer.