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

Find d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}
y=1+x1−xy=\dfrac{1+x}{1-x}

Explanation

Solution

Hint: If u and v are two differentiable functions of x then ddx(uv)=v×dudx−u×dvdxv2\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 d2ydx2\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=1+x1−xy=\dfrac{1+x}{1-x}
We have to find d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}
We know that the formula of ddx(uv)\dfrac{d}{dx}\left( \dfrac{u}{v} \right)is given by ddx(uv)=v×dudx−u×dvdxv2\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,
dydx=(1−x)(1)−(1+x)(−1)(1−x)2\dfrac{dy}{dx}=\dfrac{(1-x)\left( 1 \right)-\left( 1+x \right)\left( -1 \right)}{{{(1-x)}^{2}}}. . . . . . . . . . . . . . . . . . .(1)
dydx=1−x+1+x(1−x)2\dfrac{dy}{dx}=\dfrac{1-x+1+x}{{{(1-x)}^{2}}}
dydx=2(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.
d2ydx2=ddx(dydx)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)
d2ydx2=ddx(2(1−x)2)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{2}{{{(1-x)}^{2}}} \right). . . . . . . . . . . . . . . . . . . . . . . . . . (3)
d2ydx2=(1−x)2×0−2×2×(1−x)(−1)(1−x)4\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)
d2ydx2=2×2×(1−x)(1−x)4\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2\times 2\times (1-x)}{{{(1-x)}^{4}}} . . . . . . . . . . . . . . . . . . . . . . . . . . . (5)
d2ydx2=4(1−x)3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{4}{{{(1-x)}^{3}}}

Note:The formula for ddx(u+v)=dudx+dvdx\dfrac{d}{dx}\left( u+v \right)=\dfrac{du}{dx}+\dfrac{dv}{dx}. So the derivative of 1+x1+xis 1and this formula is known as sum rule. The formula for derivative of un=nun−1{{u}^{n}}=n{{u}^{n-1}}where n is any integer so the derivative of (1−x)2{{(1-x)}^{2}}is 2×(1−x)(−1)2\times (1-x)\left( -1 \right). Carefully do the basic mathematical operations like addition subtraction then we will get the required answer.