Find the second derivative of the following function. \[y=\dfrac{x\ln x}{1-{{x}^{2}}}\]
Answer
Verified601.5k+ views
- Hint: Use product and quotient rule to differentiate the given function twice. Also use the fact that differentiation of \[y=\ln x\] is \[\dfrac{dy}{dx}=\dfrac{1}{x}\] and differentiation of \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Complete step-by-step solution -
To find the second derivative of the function \[y=\dfrac{x\ln x}{1-{{x}^{2}}}\], 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 \[w\left( x \right)=u\left( x \right)v\left( x \right)\] then, we have \[\dfrac{d}{dx}w(x)=\dfrac{d}{dx}\left( u(x)\times v(x) \right)=v(x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}v(x)\].
Substituting \[u(x)=\dfrac{x}{1-{{x}^{2}}},v(x)=\ln x\] in the above equation, we have \[\dfrac{d}{dx}w(x)=v(x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}v(x)=\ln x\times \dfrac{d}{dx}\left( \dfrac{x}{1-{{x}^{2}}} \right)+\dfrac{x}{1-{{x}^{2}}}\times \dfrac{d}{dx}(\ln x).....\left( 1 \right)\].
We know that differentiation of function \[y=\ln x\] is \[\dfrac{dy}{dx}=\dfrac{1}{x}.....\left( 2 \right)\].
To find the value of \[\dfrac{d}{dx}\left( \dfrac{x}{1-{{x}^{2}}} \right)\], we have \[u(x)=\dfrac{x}{1-{{x}^{2}}}\].
We will use quotient rule of differentiation which states that if \[y=\dfrac{\alpha (x)}{\beta (x)}\] then, we have \[\dfrac{dy}{dx}=\dfrac{\beta (x)\alpha '(x)-\alpha (x)\beta '(x)}{{{\beta }^{2}}(x)}\].
Substituting \[\alpha (x)=x,\beta (x)=1-{{x}^{2}}\] in the above equation, we have \[\dfrac{d}{dx}u(x)=\dfrac{\beta (x)\alpha '(x)-\alpha (x)\beta '(x)}{{{\beta }^{2}}(x)}=\dfrac{(1-{{x}^{2}})\times \dfrac{d}{dx}(x)-x\times \dfrac{d}{dx}(1-{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}.....\left( 3 \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=1,b=0\] in the above equation, we have \[\dfrac{d}{dx}(x)=1.....\left( 4 \right)\].
Substituting \[a=-1,n=2,b=1\] in the above equation, we have \[\dfrac{d}{dx}(1-{{x}^{2}})=-2x.....\left( 5 \right)\].
Substituting equation (4) and (5) in equation (3), we have \[\dfrac{d}{dx}u(x)=\dfrac{(1-{{x}^{2}})\times \dfrac{d}{dx}(x)-x\times \dfrac{d}{dx}(1-{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}=\dfrac{(1-{{x}^{2}})\times 1-x\times (-2x)}{{{(1-{{x}^{2}})}^{2}}}=\dfrac{1+{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}.....\left( 6 \right)\].
Substituting equation (2) and (6) in equation (1), we get \[\dfrac{d}{dx}w(x)=\ln x\times \dfrac{d}{dx}\left( \dfrac{x}{1-{{x}^{2}}} \right)+\dfrac{x}{1-{{x}^{2}}}\times \dfrac{d}{dx}(\ln x)=\ln x\times \dfrac{1+{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{x}{1-{{x}^{2}}}\times \dfrac{1}{x}=\dfrac{\ln x(1+{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{1}{1-{{x}^{2}}}\].
Solving the above equation by taking LCM, we get \[\dfrac{d}{dx}w(x)=\dfrac{\ln x(1+{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{1}{1-{{x}^{2}}}=\dfrac{\ln x(1+{{x}^{2}})+1-{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}\].
Thus, we have \[\dfrac{dy}{dx}=\dfrac{\ln x(1+{{x}^{2}})+1-{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}\] as the first derivative of the given function.
Now, to find the second derivative, we will again differentiate the function \[\dfrac{dy}{dx}\] as \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)\].
Let’s assume \[a(x)=\dfrac{dy}{dx}=\dfrac{\ln x(1+{{x}^{2}})+1-{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}\].
So, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)\].
We can rewrite \[a(x)\] as a quotient of two functions \[\dfrac{c\left( x \right)}{d\left( x \right)}\] where \[c\left( x \right)=\ln x(1+{{x}^{2}})+1-{{x}^{2}},d\left( x \right)={{(1-{{x}^{2}})}^{2}}\].
We will use quotient rule of differentiation which states that if \[y=\dfrac{c(x)}{d(x)}\] then, we have \[\dfrac{dy}{dx}=\dfrac{d(x)c'(x)-c(x)d'(x)}{{{d}^{2}}(x)}\].
Substituting \[c\left( x \right)=\ln x(1+{{x}^{2}})+1-{{x}^{2}},d\left( x \right)={{(1-{{x}^{2}})}^{2}}\] in the above equation, we get \[\dfrac{d}{dx}a\left( x \right)=\dfrac{d(x)c'(x)-c(x)d'(x)}{{{d}^{2}}(x)}=\dfrac{{{(1-{{x}^{2}})}^{2}}\times \dfrac{d}{dx}\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\times \dfrac{dy}{dx}{{(1-{{x}^{2}})}^{2}}}{{{(1-{{x}^{2}})}^{4}}}.....\left( 7 \right)\].
To find the value of \[\dfrac{dy}{dx}{{(1-{{x}^{2}})}^{2}}\], let’s assume \[d\left( x \right)={{(1-{{x}^{2}})}^{2}}\] as a composition of two functions \[\gamma \left( \beta \left( x \right) \right)\] where \[\gamma \left( x \right)={{x}^{2}},\beta \left( x \right)=1-{{x}^{2}}\].
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 \[\gamma (x)\] and \[\beta (x)\] such that \[y=\gamma (\beta (x)))\], then \[\dfrac{dy}{dx}=\dfrac{d\gamma (\beta (x))}{d\beta (x)}\times \dfrac{d\beta (x)}{dx}\].
Substituting \[\gamma \left( x \right)={{x}^{2}},\beta \left( x \right)=1-{{x}^{2}}\] in the above equation, we get \[\dfrac{d}{dx}d\left( x \right)=\dfrac{d\gamma (\beta (x))}{d\beta (x)}\times \dfrac{d\beta (x)}{dx}=\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}\times \dfrac{d(1-{{x}^{2}})}{dx}.....\left( 8 \right)\].
To find the value of \[\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}\], let’s assume \[t=1-{{x}^{2}}\].
Thus, we have \[\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}=\dfrac{d}{dt}\left( {{t}^{2}} \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=2,b=0\] in the above equation, we get \[\dfrac{d}{dt}\left( {{t}^{2}} \right)=2t\].
So, we have \[\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}=\dfrac{d}{dt}\left( {{t}^{2}} \right)=2t=2\left( 1-{{x}^{2}} \right).....\left( 9 \right)\].
Substituting equation (5) and (9) in equation (8), we get \[\dfrac{d}{dx}d\left( x \right)=\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}\times \dfrac{d(1-{{x}^{2}})}{dx}=2\left( 1-{{x}^{2}} \right)\times \left( -2x \right)=-4x\left( 1-{{x}^{2}} \right).....\left( 10 \right)\].
To find the value of \[\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\], we can write \[c\left( x \right)=\ln x(1+{{x}^{2}})+1-{{x}^{2}}\] as a sum of two functions \[c\left( x \right)=f\left( x \right)+\beta \left( x \right)\] where \[f\left( x \right)=\ln x(1+{{x}^{2}}),\beta \left( x \right)=1-{{x}^{2}}\].
We will use sum rule of differentiation of two functions which states that if \[y=f\left( x \right)+\beta \left( x \right)\] then \[\dfrac{dy}{dx}=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}\beta \left( x \right)\].
Substituting \[f\left( x \right)=\ln x(1+{{x}^{2}}),\beta \left( x \right)=1-{{x}^{2}}\] in the above equation, we have \[\dfrac{d}{dx}c\left( x \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}\beta \left( x \right)=\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}}) \right)+\dfrac{d}{dx}\left( 1-{{x}^{2}} \right).....\left( 11 \right)\].
To find the value of \[\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}}) \right)\], we can write \[f\left( x \right)=\ln x(1+{{x}^{2}})\] as a product of two functions \[f\left( x \right)=g\left( x \right)\times h\left( x \right)\] where \[g\left( x \right)=\ln x,h\left( x \right)=1+{{x}^{2}}\].
We will use the product rule of differentiation of product of two functions which states that if \[f\left( x \right)=g\left( x \right)h\left( x \right)\] then, we have \[\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left( g(x)\times h(x) \right)=g(x)\times \dfrac{d}{dx}h(x)+h(x)\dfrac{d}{dx}g(x)\].
Substituting \[g\left( x \right)=\ln x,h\left( x \right)=1+{{x}^{2}}\] in the above equation, we have \[\dfrac{d}{dx}f(x)=g(x)\times \dfrac{d}{dx}h(x)+h(x)\dfrac{d}{dx}g(x)=\ln x\times \dfrac{d}{dx}\left( 1+{{x}^{2}} \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{d}{dx}\left( \ln x \right).....\left( 12 \right)\].
To find the value of \[\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)\], we will substitute \[a=1,n=2,b=1\] in the formula where differentiation of \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)=2x.....\left( 13 \right)\].
Substituting equation (2) and (13) in equation (12), we get \[\dfrac{d}{dx}f(x)=\ln x\times \dfrac{d}{dx}\left( 1+{{x}^{2}} \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{d}{dx}\left( \ln x \right)=\ln x\times \left( 2x \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{1}{x}\].
Solving the above equation, we get \[\dfrac{d}{dx}f(x)=\ln x\times \left( 2x \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{1}{x}=2x\ln x+\dfrac{1}{x}+x.....\left( 14 \right)\].
Substituting equation (5) and (14) in equation (11), we get \[\dfrac{d}{dx}c\left( x \right)=\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}}) \right)+\dfrac{d}{dx}\left( 1-{{x}^{2}} \right)=2x\ln x+\dfrac{1}{x}+x-2x=2x\ln x+\dfrac{1}{x}-x.....\left( 15 \right)\].
Substituting equation (10) and (15) in equation (7), we get \[\dfrac{d}{dx}a\left( x \right)=\dfrac{{{(1-{{x}^{2}})}^{2}}\times \dfrac{d}{dx}\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\times \dfrac{dy}{dx}{{(1-{{x}^{2}})}^{2}}}{{{(1-{{x}^{2}})}^{4}}}=\dfrac{{{(1-{{x}^{2}})}^{2}}\left( 2x\ln x+\dfrac{1}{x}-x \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\left( -4x\left( 1-{{x}^{2}} \right) \right)}{{{(1-{{x}^{2}})}^{4}}}\].
Solving the above equation, we get \[\dfrac{d}{dx}a\left( x \right)=\dfrac{{{(1-{{x}^{2}})}^{2}}\left( 2x\ln x+\dfrac{1}{x}-x \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\left( -4x\left( 1-{{x}^{2}} \right) \right)}{{{(1-{{x}^{2}})}^{4}}}=\dfrac{2x\ln x+\dfrac{1}{x}-x}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{4x\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)}{{{(1-{{x}^{2}})}^{3}}}\].
Thus, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)=\dfrac{2x\ln x+\dfrac{1}{x}-x}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{4x\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)}{{{(1-{{x}^{2}})}^{3}}}\].
Hence, the second derivative of \[y=\dfrac{x\ln x}{1-{{x}^{2}}}\] is \[\dfrac{2x\ln x+\dfrac{1}{x}-x}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{4x\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)}{{{(1-{{x}^{2}})}^{3}}}\].
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.
Complete step-by-step solution -
To find the second derivative of the function \[y=\dfrac{x\ln x}{1-{{x}^{2}}}\], 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 \[w\left( x \right)=u\left( x \right)v\left( x \right)\] then, we have \[\dfrac{d}{dx}w(x)=\dfrac{d}{dx}\left( u(x)\times v(x) \right)=v(x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}v(x)\].
Substituting \[u(x)=\dfrac{x}{1-{{x}^{2}}},v(x)=\ln x\] in the above equation, we have \[\dfrac{d}{dx}w(x)=v(x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}v(x)=\ln x\times \dfrac{d}{dx}\left( \dfrac{x}{1-{{x}^{2}}} \right)+\dfrac{x}{1-{{x}^{2}}}\times \dfrac{d}{dx}(\ln x).....\left( 1 \right)\].
We know that differentiation of function \[y=\ln x\] is \[\dfrac{dy}{dx}=\dfrac{1}{x}.....\left( 2 \right)\].
To find the value of \[\dfrac{d}{dx}\left( \dfrac{x}{1-{{x}^{2}}} \right)\], we have \[u(x)=\dfrac{x}{1-{{x}^{2}}}\].
We will use quotient rule of differentiation which states that if \[y=\dfrac{\alpha (x)}{\beta (x)}\] then, we have \[\dfrac{dy}{dx}=\dfrac{\beta (x)\alpha '(x)-\alpha (x)\beta '(x)}{{{\beta }^{2}}(x)}\].
Substituting \[\alpha (x)=x,\beta (x)=1-{{x}^{2}}\] in the above equation, we have \[\dfrac{d}{dx}u(x)=\dfrac{\beta (x)\alpha '(x)-\alpha (x)\beta '(x)}{{{\beta }^{2}}(x)}=\dfrac{(1-{{x}^{2}})\times \dfrac{d}{dx}(x)-x\times \dfrac{d}{dx}(1-{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}.....\left( 3 \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=1,b=0\] in the above equation, we have \[\dfrac{d}{dx}(x)=1.....\left( 4 \right)\].
Substituting \[a=-1,n=2,b=1\] in the above equation, we have \[\dfrac{d}{dx}(1-{{x}^{2}})=-2x.....\left( 5 \right)\].
Substituting equation (4) and (5) in equation (3), we have \[\dfrac{d}{dx}u(x)=\dfrac{(1-{{x}^{2}})\times \dfrac{d}{dx}(x)-x\times \dfrac{d}{dx}(1-{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}=\dfrac{(1-{{x}^{2}})\times 1-x\times (-2x)}{{{(1-{{x}^{2}})}^{2}}}=\dfrac{1+{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}.....\left( 6 \right)\].
Substituting equation (2) and (6) in equation (1), we get \[\dfrac{d}{dx}w(x)=\ln x\times \dfrac{d}{dx}\left( \dfrac{x}{1-{{x}^{2}}} \right)+\dfrac{x}{1-{{x}^{2}}}\times \dfrac{d}{dx}(\ln x)=\ln x\times \dfrac{1+{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{x}{1-{{x}^{2}}}\times \dfrac{1}{x}=\dfrac{\ln x(1+{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{1}{1-{{x}^{2}}}\].
Solving the above equation by taking LCM, we get \[\dfrac{d}{dx}w(x)=\dfrac{\ln x(1+{{x}^{2}})}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{1}{1-{{x}^{2}}}=\dfrac{\ln x(1+{{x}^{2}})+1-{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}\].
Thus, we have \[\dfrac{dy}{dx}=\dfrac{\ln x(1+{{x}^{2}})+1-{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}\] as the first derivative of the given function.
Now, to find the second derivative, we will again differentiate the function \[\dfrac{dy}{dx}\] as \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)\].
Let’s assume \[a(x)=\dfrac{dy}{dx}=\dfrac{\ln x(1+{{x}^{2}})+1-{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}}\].
So, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)\].
We can rewrite \[a(x)\] as a quotient of two functions \[\dfrac{c\left( x \right)}{d\left( x \right)}\] where \[c\left( x \right)=\ln x(1+{{x}^{2}})+1-{{x}^{2}},d\left( x \right)={{(1-{{x}^{2}})}^{2}}\].
We will use quotient rule of differentiation which states that if \[y=\dfrac{c(x)}{d(x)}\] then, we have \[\dfrac{dy}{dx}=\dfrac{d(x)c'(x)-c(x)d'(x)}{{{d}^{2}}(x)}\].
Substituting \[c\left( x \right)=\ln x(1+{{x}^{2}})+1-{{x}^{2}},d\left( x \right)={{(1-{{x}^{2}})}^{2}}\] in the above equation, we get \[\dfrac{d}{dx}a\left( x \right)=\dfrac{d(x)c'(x)-c(x)d'(x)}{{{d}^{2}}(x)}=\dfrac{{{(1-{{x}^{2}})}^{2}}\times \dfrac{d}{dx}\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\times \dfrac{dy}{dx}{{(1-{{x}^{2}})}^{2}}}{{{(1-{{x}^{2}})}^{4}}}.....\left( 7 \right)\].
To find the value of \[\dfrac{dy}{dx}{{(1-{{x}^{2}})}^{2}}\], let’s assume \[d\left( x \right)={{(1-{{x}^{2}})}^{2}}\] as a composition of two functions \[\gamma \left( \beta \left( x \right) \right)\] where \[\gamma \left( x \right)={{x}^{2}},\beta \left( x \right)=1-{{x}^{2}}\].
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 \[\gamma (x)\] and \[\beta (x)\] such that \[y=\gamma (\beta (x)))\], then \[\dfrac{dy}{dx}=\dfrac{d\gamma (\beta (x))}{d\beta (x)}\times \dfrac{d\beta (x)}{dx}\].
Substituting \[\gamma \left( x \right)={{x}^{2}},\beta \left( x \right)=1-{{x}^{2}}\] in the above equation, we get \[\dfrac{d}{dx}d\left( x \right)=\dfrac{d\gamma (\beta (x))}{d\beta (x)}\times \dfrac{d\beta (x)}{dx}=\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}\times \dfrac{d(1-{{x}^{2}})}{dx}.....\left( 8 \right)\].
To find the value of \[\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}\], let’s assume \[t=1-{{x}^{2}}\].
Thus, we have \[\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}=\dfrac{d}{dt}\left( {{t}^{2}} \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=2,b=0\] in the above equation, we get \[\dfrac{d}{dt}\left( {{t}^{2}} \right)=2t\].
So, we have \[\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}=\dfrac{d}{dt}\left( {{t}^{2}} \right)=2t=2\left( 1-{{x}^{2}} \right).....\left( 9 \right)\].
Substituting equation (5) and (9) in equation (8), we get \[\dfrac{d}{dx}d\left( x \right)=\dfrac{d{{(1-{{x}^{2}})}^{2}}}{d(1-{{x}^{2}})}\times \dfrac{d(1-{{x}^{2}})}{dx}=2\left( 1-{{x}^{2}} \right)\times \left( -2x \right)=-4x\left( 1-{{x}^{2}} \right).....\left( 10 \right)\].
To find the value of \[\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\], we can write \[c\left( x \right)=\ln x(1+{{x}^{2}})+1-{{x}^{2}}\] as a sum of two functions \[c\left( x \right)=f\left( x \right)+\beta \left( x \right)\] where \[f\left( x \right)=\ln x(1+{{x}^{2}}),\beta \left( x \right)=1-{{x}^{2}}\].
We will use sum rule of differentiation of two functions which states that if \[y=f\left( x \right)+\beta \left( x \right)\] then \[\dfrac{dy}{dx}=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}\beta \left( x \right)\].
Substituting \[f\left( x \right)=\ln x(1+{{x}^{2}}),\beta \left( x \right)=1-{{x}^{2}}\] in the above equation, we have \[\dfrac{d}{dx}c\left( x \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}\beta \left( x \right)=\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}}) \right)+\dfrac{d}{dx}\left( 1-{{x}^{2}} \right).....\left( 11 \right)\].
To find the value of \[\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}}) \right)\], we can write \[f\left( x \right)=\ln x(1+{{x}^{2}})\] as a product of two functions \[f\left( x \right)=g\left( x \right)\times h\left( x \right)\] where \[g\left( x \right)=\ln x,h\left( x \right)=1+{{x}^{2}}\].
We will use the product rule of differentiation of product of two functions which states that if \[f\left( x \right)=g\left( x \right)h\left( x \right)\] then, we have \[\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left( g(x)\times h(x) \right)=g(x)\times \dfrac{d}{dx}h(x)+h(x)\dfrac{d}{dx}g(x)\].
Substituting \[g\left( x \right)=\ln x,h\left( x \right)=1+{{x}^{2}}\] in the above equation, we have \[\dfrac{d}{dx}f(x)=g(x)\times \dfrac{d}{dx}h(x)+h(x)\dfrac{d}{dx}g(x)=\ln x\times \dfrac{d}{dx}\left( 1+{{x}^{2}} \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{d}{dx}\left( \ln x \right).....\left( 12 \right)\].
To find the value of \[\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)\], we will substitute \[a=1,n=2,b=1\] in the formula where differentiation of \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Thus, we have \[\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)=2x.....\left( 13 \right)\].
Substituting equation (2) and (13) in equation (12), we get \[\dfrac{d}{dx}f(x)=\ln x\times \dfrac{d}{dx}\left( 1+{{x}^{2}} \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{d}{dx}\left( \ln x \right)=\ln x\times \left( 2x \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{1}{x}\].
Solving the above equation, we get \[\dfrac{d}{dx}f(x)=\ln x\times \left( 2x \right)+\left( 1+{{x}^{2}} \right)\times \dfrac{1}{x}=2x\ln x+\dfrac{1}{x}+x.....\left( 14 \right)\].
Substituting equation (5) and (14) in equation (11), we get \[\dfrac{d}{dx}c\left( x \right)=\dfrac{d}{dx}\left( \ln x(1+{{x}^{2}}) \right)+\dfrac{d}{dx}\left( 1-{{x}^{2}} \right)=2x\ln x+\dfrac{1}{x}+x-2x=2x\ln x+\dfrac{1}{x}-x.....\left( 15 \right)\].
Substituting equation (10) and (15) in equation (7), we get \[\dfrac{d}{dx}a\left( x \right)=\dfrac{{{(1-{{x}^{2}})}^{2}}\times \dfrac{d}{dx}\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\times \dfrac{dy}{dx}{{(1-{{x}^{2}})}^{2}}}{{{(1-{{x}^{2}})}^{4}}}=\dfrac{{{(1-{{x}^{2}})}^{2}}\left( 2x\ln x+\dfrac{1}{x}-x \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\left( -4x\left( 1-{{x}^{2}} \right) \right)}{{{(1-{{x}^{2}})}^{4}}}\].
Solving the above equation, we get \[\dfrac{d}{dx}a\left( x \right)=\dfrac{{{(1-{{x}^{2}})}^{2}}\left( 2x\ln x+\dfrac{1}{x}-x \right)-\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)\left( -4x\left( 1-{{x}^{2}} \right) \right)}{{{(1-{{x}^{2}})}^{4}}}=\dfrac{2x\ln x+\dfrac{1}{x}-x}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{4x\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)}{{{(1-{{x}^{2}})}^{3}}}\].
Thus, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)=\dfrac{2x\ln x+\dfrac{1}{x}-x}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{4x\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)}{{{(1-{{x}^{2}})}^{3}}}\].
Hence, the second derivative of \[y=\dfrac{x\ln x}{1-{{x}^{2}}}\] is \[\dfrac{2x\ln x+\dfrac{1}{x}-x}{{{(1-{{x}^{2}})}^{2}}}+\dfrac{4x\left( \ln x(1+{{x}^{2}})+1-{{x}^{2}} \right)}{{{(1-{{x}^{2}})}^{3}}}\].
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.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success
Why are manures considered better than fertilizers class 11 biology CBSE
Find the coordinates of the midpoint of the line segment class 11 maths CBSE
Distinguish between static friction limiting friction class 11 physics CBSE
The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE
The first National Commission on Labour NCL submitted class 11 social science CBSE
Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE
10 examples of friction in our daily life
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
What are Quantum numbers Explain the quantum number class 11 chemistry CBSE
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE