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Question: Find the second derivative of the following function. \[y=\dfrac{x\ln x}{1-{{x}^{2}}}\]...

Find the second derivative of the following function. y=xlnx1x2y=\dfrac{x\ln x}{1-{{x}^{2}}}

Explanation

Solution

- Hint: Use product and quotient rule to differentiate the given function twice. Also use the fact that differentiation of y=lnxy=\ln x is dydx=1x\dfrac{dy}{dx}=\dfrac{1}{x} and differentiation of y=axn+by=a{{x}^{n}}+b is dydx=anxn1\dfrac{dy}{dx}=an{{x}^{n-1}}.

Complete step-by-step solution -

To find the second derivative of the function y=xlnx1x2y=\dfrac{x\ln x}{1-{{x}^{2}}}, we will differentiate it twice with respect to the variable x.
To find the first derivative, we will differentiate the function y with respect to x.
We will use the product rule of differentiation of product of two functions which states that if w(x)=u(x)v(x)w\left( x \right)=u\left( x \right)v\left( x \right) then, we have ddxw(x)=ddx(u(x)×v(x))=v(x)×ddxu(x)+u(x)ddxv(x)\dfrac{d}{dx}w(x)=\dfrac{d}{dx}\left( u(x)\times v(x) \right)=v(x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}v(x).
Substituting u(x)=x1x2,v(x)=lnxu(x)=\dfrac{x}{1-{{x}^{2}}},v(x)=\ln x in the above equation, we have ddxw(x)=v(x)×ddxu(x)+u(x)ddxv(x)=lnx×ddx(x1x2)+x1x2×ddx(lnx).....(1)\dfrac{d}{dx}w(x)=v(x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}v(x)=\ln x\times \dfrac{d}{dx}\left( \dfrac{x}{1-{{x}^{2}}} \right)+\dfrac{x}{1-{{x}^{2}}}\times \dfrac{d}{dx}(\ln x).....\left( 1 \right).
We know that differentiation of function y=lnxy=\ln x is dydx=1x.....(2)\dfrac{dy}{dx}=\dfrac{1}{x}.....\left( 2 \right).
To find the value of ddx(x1x2)\dfrac{d}{dx}\left( \dfrac{x}{1-{{x}^{2}}} \right), we have u(x)=x1x2u(x)=\dfrac{x}{1-{{x}^{2}}}.
We will use quotient rule of differentiation which states that if y=α(x)β(x)y=\dfrac{\alpha (x)}{\beta (x)} then, we have dydx=β(x)α(x)α(x)β(x)β2(x)\dfrac{dy}{dx}=\dfrac{\beta (x)\alpha '(x)-\alpha (x)\beta '(x)}{{{\beta }^{2}}(x)}.
Substituting α(x)=x,β(x)=1x2\alpha (x)=x,\beta (x)=1-{{x}^{2}} in the above equation, we have ddxu(x)=β(x)α(x)α(x)β(x)β2(x)=(1x2)×ddx(x)x×ddx(1x2)(1x2)2.....(3)\dfrac{d}{dx}u(x)=\dfrac{\beta (x)\alpha '(x)-\alpha (x)\beta '(x)}{{{\beta }^{2}}(x)}=\dfrac{(1-{{x}^{2}})\times \dfrac{d}{dx}(x)-x\times \dfrac{d}{dx}(1-{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}.....\left( 3 \right).
We know that differentiation of any function of the form y=axn+by=a{{x}^{n}}+b is dydx=anxn1\dfrac{dy}{dx}=an{{x}^{n-1}}.
Substituting a=1,n=1,b=0a=1,n=1,b=0 in the above equation, we have ddx(x)=1.....(4)\dfrac{d}{dx}(x)=1.....\left( 4 \right).
Substituting a=1,n=2,b=1a=-1,n=2,b=1 in the above equation, we have ddx(1x2)=2x.....(5)\dfrac{d}{dx}(1-{{x}^{2}})=-2x.....\left( 5 \right).
Substituting equation (4) and (5) in equation (3), we have ddxu(x)=(1x2)×ddx(x)x×ddx(1x2)(1x2)2=(1x2)×1x×(2x)(1x2)2=1+x2(1x2)2.....(6)\dfrac{d}{dx}u(x)=\dfrac{(1-{{x}^{2}})\times \dfrac{d}{dx}(x)-x\times \dfrac{d}{dx}(1-{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}=\dfrac{(1-{{x}^{2}})\times 1-x\times (-2x)}{{{(1-{{x}^{2}})}^{2}}}=\dfrac{1+{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}.....\left( 6 \right).
Substituting equation (2) and (6) in equation (1), we get ddxw(x)=lnx×ddx(x1x2)+x1x2×ddx(lnx)=lnx×1+x2(1x2)2+x1x2×1x=lnx(1+x2)(1x2)2+11x2\dfrac{d}{dx}w(x)=\ln x\times \dfrac{d}{dx}\left( \dfrac{x}{1-{{x}^{2}}} \right)+\dfrac{x}{1-{{x}^{2}}}\times \dfrac{d}{dx}(\ln x)=\ln x\times \dfrac{1+{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{x}{1-{{x}^{2}}}\times \dfrac{1}{x}=\dfrac{\ln x(1+{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{1}{1-{{x}^{2}}}.
Solving the above equation by taking LCM, we get ddxw(x)=lnx(1+x2)(1x2)2+11x2=lnx(1+x2)+1x2(1x2)2\dfrac{d}{dx}w(x)=\dfrac{\ln x(1+{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{1}{1-{{x}^{2}}}=\dfrac{\ln x(1+{{x}^{2}})+1-{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}.
Thus, we have dydx=lnx(1+x2)+1x2(1x2)2\dfrac{dy}{dx}=\dfrac{\ln x(1+{{x}^{2}})+1-{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}} as the first derivative of the given function.
Now, to find the second derivative, we will again differentiate the function dydx\dfrac{dy}{dx} as d2ydx2=ddx(dydx)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right).
Let’s assume a(x)=dydx=lnx(1+x2)+1x2(1x2)2a(x)=\dfrac{dy}{dx}=\dfrac{\ln x(1+{{x}^{2}})+1-{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}.
So, we have d2ydx2=ddxa(x)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x).
We can rewrite a(x)a(x) as a quotient of two functions c(x)d(x)\dfrac{c\left( x \right)}{d\left( x \right)} where c(x)=lnx(1+x2)+1x2,d(x)=(1x2)2c\left( x \right)=\ln x(1+{{x}^{2}})+1-{{x}^{2}},d\left( x \right)={{(1-{{x}^{2}})}^{2}}.
We will use quotient rule of differentiation which states that if y=c(x)d(x)y=\dfrac{c(x)}{d(x)} then, we have dydx=d(x)c(x)c(x)d(x)d2(x)\dfrac{dy}{dx}=\dfrac{d(x)c'(x)-c(x)d'(x)}{{{d}^{2}}(x)}.
Substituting c(x)=lnx(1+x2)+1x2,d(x)=(1x2)2c\left( x \right)=\ln x(1+{{x}^{2}})+1-{{x}^{2}},d\left( x \right)={{(1-{{x}^{2}})}^{2}} in the above equation, we get ddxa(x)=d(x)c(x)c(x)d(x)d2(x)=(1x2)2×ddx(lnx(1+x2)+1x2)(lnx(1+x2)+1x2)×dydx(1x2)2(1x2)4.....(7)\dfrac{d}{dx}a\left( x \right)=\dfrac{d(x)c'(x)-c(x)d'(x)}{{{d}^{2}}(x)}=\dfrac{{{(1-{{x}^{2}})}^{2}}\times \dfrac{d}{dx}\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\times \dfrac{dy}{dx}{{(1-{{x}^{2}})}^{2}}}{{{(1-{{x}^{2}})}^{4}}}.....\left( 7 \right).
To find the value of dydx(1x2)2\dfrac{dy}{dx}{{(1-{{x}^{2}})}^{2}}, let’s assume d(x)=(1x2)2d\left( x \right)={{(1-{{x}^{2}})}^{2}} as a composition of two functions γ(β(x))\gamma \left( \beta \left( x \right) \right) where γ(x)=x2,β(x)=1x2\gamma \left( x \right)={{x}^{2}},\beta \left( x \right)=1-{{x}^{2}}.
We will use chain rule of differentiation of composition of two functions to find the derivative which states that if y is a composition of two functions γ(x)\gamma (x) and β(x)\beta (x) such that y=γ(β(x)))y=\gamma (\beta (x))), then dydx=dγ(β(x))dβ(x)×dβ(x)dx\dfrac{dy}{dx}=\dfrac{d\gamma (\beta (x))}{d\beta (x)}\times \dfrac{d\beta (x)}{dx}.
Substituting γ(x)=x2,β(x)=1x2\gamma \left( x \right)={{x}^{2}},\beta \left( x \right)=1-{{x}^{2}} in the above equation, we get ddxd(x)=dγ(β(x))dβ(x)×dβ(x)dx=d(1x2)2d(1x2)×d(1x2)dx.....(8)\dfrac{d}{dx}d\left( x \right)=\dfrac{d\gamma (\beta (x))}{d\beta (x)}\times \dfrac{d\beta (x)}{dx}=\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}\times \dfrac{d(1-{{x}^{2}})}{dx}.....\left( 8 \right).
To find the value of d(1x2)2d(1x2)\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}, let’s assume t=1x2t=1-{{x}^{2}}.
Thus, we have d(1x2)2d(1x2)=ddt(t2)\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}=\dfrac{d}{dt}\left( {{t}^{2}} \right).
We know that differentiation of any function of the form y=axn+by=a{{x}^{n}}+b is dydx=anxn1\dfrac{dy}{dx}=an{{x}^{n-1}}.
Substituting a=1,n=2,b=0a=1,n=2,b=0 in the above equation, we get ddt(t2)=2t\dfrac{d}{dt}\left( {{t}^{2}} \right)=2t.
So, we have d(1x2)2d(1x2)=ddt(t2)=2t=2(1x2).....(9)\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}=\dfrac{d}{dt}\left( {{t}^{2}} \right)=2t=2\left( 1-{{x}^{2}} \right).....\left( 9 \right).
Substituting equation (5) and (9) in equation (8), we get ddxd(x)=d(1x2)2d(1x2)×d(1x2)dx=2(1x2)×(2x)=4x(1x2).....(10)\dfrac{d}{dx}d\left( x \right)=\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}\times \dfrac{d(1-{{x}^{2}})}{dx}=2\left( 1-{{x}^{2}} \right)\times \left( -2x \right)=-4x\left( 1-{{x}^{2}} \right).....\left( 10 \right).
To find the value of ddx(lnx(1+x2)+1x2)\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right), we can write c(x)=lnx(1+x2)+1x2c\left( x \right)=\ln x(1+{{x}^{2}})+1-{{x}^{2}} as a sum of two functions c(x)=f(x)+β(x)c\left( x \right)=f\left( x \right)+\beta \left( x \right) where f(x)=lnx(1+x2),β(x)=1x2f\left( x \right)=\ln x(1+{{x}^{2}}),\beta \left( x \right)=1-{{x}^{2}}.
We will use sum rule of differentiation of two functions which states that if y=f(x)+β(x)y=f\left( x \right)+\beta \left( x \right) then dydx=ddxf(x)+ddxβ(x)\dfrac{dy}{dx}=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}\beta \left( x \right).
Substituting f(x)=lnx(1+x2),β(x)=1x2f\left( x \right)=\ln x(1+{{x}^{2}}),\beta \left( x \right)=1-{{x}^{2}} in the above equation, we have ddxc(x)=ddxf(x)+ddxβ(x)=ddx(lnx(1+x2))+ddx(1x2).....(11)\dfrac{d}{dx}c\left( x \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}\beta \left( x \right)=\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}}) \right)+\dfrac{d}{dx}\left( 1-{{x}^{2}} \right).....\left( 11 \right).
To find the value of ddx(lnx(1+x2))\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}}) \right), we can write f(x)=lnx(1+x2)f\left( x \right)=\ln x(1+{{x}^{2}}) as a product of two functions f(x)=g(x)×h(x)f\left( x \right)=g\left( x \right)\times h\left( x \right) where g(x)=lnx,h(x)=1+x2g\left( x \right)=\ln x,h\left( x \right)=1+{{x}^{2}}.
We will use the product rule of differentiation of product of two functions which states that if f(x)=g(x)h(x)f\left( x \right)=g\left( x \right)h\left( x \right) then, we have ddxf(x)=ddx(g(x)×h(x))=g(x)×ddxh(x)+h(x)ddxg(x)\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left( g(x)\times h(x) \right)=g(x)\times \dfrac{d}{dx}h(x)+h(x)\dfrac{d}{dx}g(x).
Substituting g(x)=lnx,h(x)=1+x2g\left( x \right)=\ln x,h\left( x \right)=1+{{x}^{2}} in the above equation, we have ddxf(x)=g(x)×ddxh(x)+h(x)ddxg(x)=lnx×ddx(1+x2)+(1+x2)×ddx(lnx).....(12)\dfrac{d}{dx}f(x)=g(x)\times \dfrac{d}{dx}h(x)+h(x)\dfrac{d}{dx}g(x)=\ln x\times \dfrac{d}{dx}\left( 1+{{x}^{2}} \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{d}{dx}\left( \ln x \right).....\left( 12 \right).
To find the value of ddx(1+x2)\dfrac{d}{dx}\left( 1+{{x}^{2}} \right), we will substitute a=1,n=2,b=1a=1,n=2,b=1 in the formula where differentiation of y=axn+by=a{{x}^{n}}+b is dydx=anxn1\dfrac{dy}{dx}=an{{x}^{n-1}}.
Thus, we have ddx(1+x2)=2x.....(13)\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)=2x.....\left( 13 \right).
Substituting equation (2) and (13) in equation (12), we get ddxf(x)=lnx×ddx(1+x2)+(1+x2)×ddx(lnx)=lnx×(2x)+(1+x2)×1x\dfrac{d}{dx}f(x)=\ln x\times \dfrac{d}{dx}\left( 1+{{x}^{2}} \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{d}{dx}\left( \ln x \right)=\ln x\times \left( 2x \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{1}{x}.
Solving the above equation, we get ddxf(x)=lnx×(2x)+(1+x2)×1x=2xlnx+1x+x.....(14)\dfrac{d}{dx}f(x)=\ln x\times \left( 2x \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{1}{x}=2x\ln x+\dfrac{1}{x}+x.....\left( 14 \right).
Substituting equation (5) and (14) in equation (11), we get ddxc(x)=ddx(lnx(1+x2))+ddx(1x2)=2xlnx+1x+x2x=2xlnx+1xx.....(15)\dfrac{d}{dx}c\left( x \right)=\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}}) \right)+\dfrac{d}{dx}\left( 1-{{x}^{2}} \right)=2x\ln x+\dfrac{1}{x}+x-2x=2x\ln x+\dfrac{1}{x}-x.....\left( 15 \right).
Substituting equation (10) and (15) in equation (7), we get ddxa(x)=(1x2)2×ddx(lnx(1+x2)+1x2)(lnx(1+x2)+1x2)×dydx(1x2)2(1x2)4=(1x2)2(2xlnx+1xx)(lnx(1+x2)+1x2)(4x(1x2))(1x2)4\dfrac{d}{dx}a\left( x \right)=\dfrac{{{(1-{{x}^{2}})}^{2}}\times \dfrac{d}{dx}\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\times \dfrac{dy}{dx}{{(1-{{x}^{2}})}^{2}}}{{{(1-{{x}^{2}})}^{4}}}=\dfrac{{{(1-{{x}^{2}})}^{2}}\left( 2x\ln x+\dfrac{1}{x}-x \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\left( -4x\left( 1-{{x}^{2}} \right) \right)}{{{(1-{{x}^{2}})}^{4}}}.
Solving the above equation, we get ddxa(x)=(1x2)2(2xlnx+1xx)(lnx(1+x2)+1x2)(4x(1x2))(1x2)4=2xlnx+1xx(1x2)2+4x(lnx(1+x2)+1x2)(1x2)3\dfrac{d}{dx}a\left( x \right)=\dfrac{{{(1-{{x}^{2}})}^{2}}\left( 2x\ln x+\dfrac{1}{x}-x \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\left( -4x\left( 1-{{x}^{2}} \right) \right)}{{{(1-{{x}^{2}})}^{4}}}=\dfrac{2x\ln x+\dfrac{1}{x}-x}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{4x\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)}{{{(1-{{x}^{2}})}^{3}}}.
Thus, we have d2ydx2=ddxa(x)=2xlnx+1xx(1x2)2+4x(lnx(1+x2)+1x2)(1x2)3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)=\dfrac{2x\ln x+\dfrac{1}{x}-x}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{4x\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)}{{{(1-{{x}^{2}})}^{3}}}.
Hence, the second derivative of y=xlnx1x2y=\dfrac{x\ln x}{1-{{x}^{2}}} is 2xlnx+1xx(1x2)2+4x(lnx(1+x2)+1x2)(1x2)3\dfrac{2x\ln x+\dfrac{1}{x}-x}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{4x\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)}{{{(1-{{x}^{2}})}^{3}}}.

Note: It is necessary to use the chain rule of differentiation for composition of two functions. Otherwise, we won’t be able to find the second derivative of the given function. Also, one must know that the first derivative of any function is the slope of the function.