Find the second derivative of the following function. \[y=x\arcsin x\]
Answer
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- Hint: One must know that \[\arcsin x={{\sin }^{-1}}x\].Use product rule of two functions to differentiate the given function twice. Also use the fact that differentiation of \[y=\arcsin x\] is \[\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\] 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 \[y=x\arcsin x\], 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 \[\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left( g(x)\times h(x) \right)=g(x)\times \dfrac{d}{dx}h(x)+h(x)\dfrac{d}{dx}g(x)\].
Substituting \[g(x)=x,h(x)=\arcsin x\] in the above equation, we have \[\dfrac{df(x)}{dx}=\dfrac{d}{dx}(x\arcsin x)=x\times \dfrac{d}{dx}(\arcsin x)+\arcsin x\times \dfrac{d}{dx}(x).....\left( 1 \right)\] .
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 \[\dfrac{d}{dx}g(x)=\dfrac{d}{dx}(x)=1.....\left( 2 \right)\].
We know that differentiation of any function of the form \[y=\arcsin x\] is \[\dfrac{d}{dx}h(x)=\dfrac{d}{dx}(\arcsin x)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}.....\left( 3 \right)\].
Substituting the value of equation (2) and (3) in equation (1), we get \[\dfrac{df(x)}{dx}=x\times \dfrac{d}{dx}(\arcsin x)+\arcsin x\times \dfrac{d}{dx}(x)=x\times \dfrac{1}{\sqrt{1-{{x}^{2}}}}+\arcsin x=\dfrac{x}{\sqrt{1-{{x}^{2}}}}+\arcsin x.....\left( 4 \right)\].
Thus, we have \[\dfrac{dy}{dx}=\dfrac{x}{\sqrt{1-{{x}^{2}}}}+\arcsin x\] 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 \[a(x)=\dfrac{dy}{dx}=\dfrac{x}{\sqrt{1-{{x}^{2}}}}+\arcsin x\].
So, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)\].
We can write \[a(x)\] as a sum of two functions \[a(x)=u(x)+v(x)\].
We know that the sum rule for differentiation of two functions is such that if \[y=a(x)=u(x)+v(x)\] then \[\dfrac{dy}{dx}=\dfrac{d}{dx}a(x)=\dfrac{d}{dx}u(x)+\dfrac{d}{dx}v(x)\].
Substituting \[u(x)=\dfrac{x}{\sqrt{1-{{x}^{2}}}},v(x)=\arcsin x\] in the above equation, we have \[\dfrac{dy}{dx}=\dfrac{d}{dx}a(x)=\dfrac{d}{dx}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)+\dfrac{d}{dx}(\arcsin x).....\left( 5 \right)\].
To find the value of \[\dfrac{d}{dx}u(x)\], 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 \[\alpha (x)=x,\beta (x)=\sqrt{1-{{x}^{2}}}\] in the above equation, we get \[\dfrac{d}{dx}u(x)=\dfrac{\sqrt{1-{{x}^{2}}}\times \dfrac{d}{dx}(x)-x\times \dfrac{d}{dx}(\sqrt{1-{{x}^{2}}})}{{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}.....\left( 6 \right)\].
Let’s assume \[t(x)=\sqrt{1-{{x}^{2}}}\].
We can write \[t(x)\] as a composition of two functions such that \[t(x)=w(z(x))\] where \[w(x)=\sqrt{x,}z(x)=1-{{x}^{2}}\].
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 \[w(x)\] and \[z(x)\] such that \[y=w(z(x))\], then \[\dfrac{dy}{dx}=\dfrac{dw(z(x))}{dz(x)}\times \dfrac{dz(x)}{dx}\].
Substituting \[w(x)=\sqrt{x,}z(x)=1-{{x}^{2}}\] in the above equation, we get \[\dfrac{dy}{dx}=\dfrac{d}{dx}t(x)=\dfrac{dw(z(x))}{dz(x)}\times \dfrac{d}{dx}z(x)=\dfrac{d(\sqrt{1-{{x}^{2}}})}{d(1-{{x}^{2}})}\times \dfrac{d}{dx}(1-{{x}^{2}}).....\left( 7 \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=-1,n=2,b=1\] in the above equation, we get \[\dfrac{d}{dx}(1-{{x}^{2}})=-2x.....\left( 8 \right)\].
To find the value of \[\dfrac{d(\sqrt{1-{{x}^{2}}})}{d(1-{{x}^{2}})}\], let’s assume \[\gamma =1-{{x}^{2}}\].
Thus, we have \[\dfrac{d(\sqrt{1-{{x}^{2}}})}{d(1-{{x}^{2}})}=\dfrac{d}{d\gamma }(\sqrt{\gamma })\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=\dfrac{1}{2},b=0\] in the above equation, we have \[\dfrac{d}{d\gamma }(\sqrt{\gamma })=\dfrac{1}{2\sqrt{\gamma }}\].
So, we have \[\dfrac{d(\sqrt{1-{{x}^{2}}})}{d(1-{{x}^{2}})}=\dfrac{d}{d\gamma }(\sqrt{\gamma })=\dfrac{1}{2\sqrt{\gamma }}=\dfrac{1}{2\sqrt{1-{{x}^{2}}}}.....\left( 9 \right)\].
Substituting equation (8) and (9) in equation (7), we get \[\dfrac{dy}{dx}=\dfrac{d}{dx}t(x)=\dfrac{d(\sqrt{1-{{x}^{2}}})}{d(1-{{x}^{2}})}\times \dfrac{d}{dx}(1-{{x}^{2}})=\dfrac{1}{2\sqrt{1-{{x}^{2}}}}\times (-2x)=\dfrac{-x}{\sqrt{1-{{x}^{2}}}}.....\left( 10 \right)\].
Substituting equation (2) and (10) in equation (6), we get \[\dfrac{d}{dx}u(x)=\dfrac{\sqrt{1-{{x}^{2}}}\times \dfrac{d}{dx}(x)-x\times \dfrac{d}{dx}(\sqrt{1-{{x}^{2}}})}{{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}=\dfrac{\sqrt{1-{{x}^{2}}}\times 1-x\times \left( \dfrac{-x}{\sqrt{1-{{x}^{2}}}} \right)}{1-{{x}^{2}}}\].
Solving the above equation by taking LCM, we have \[\dfrac{d}{dx}u(x)=\dfrac{\sqrt{1-{{x}^{2}}}\times 1-x\times \left( \dfrac{-x}{\sqrt{1-{{x}^{2}}}} \right)}{1-{{x}^{2}}}=\dfrac{(1-{{x}^{2}})+{{x}^{2}}}{{{(1-{{x}^{2}})}^{\dfrac{3}{2}}}}=\dfrac{1}{{{(1-{{x}^{2}})}^{\dfrac{3}{2}}}}={{(1-{{x}^{2}})}^{\dfrac{-3}{2}}}.....\left( 11 \right)\].
Substituting equation (3) and (11) in equation (5) and solving by taking LCM, we have \[\dfrac{d}{dx}a(x)=\dfrac{d}{dx}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)+\dfrac{d}{dx}(\arcsin x)={{(1-{{x}^{2}})}^{\dfrac{-3}{2}}}+\dfrac{1}{\sqrt{1-{{x}^{2}}}}=\dfrac{1+(1-{{x}^{2}})}{{{(1-{{x}^{2}})}^{\dfrac{3}{2}}}}=\dfrac{2-{{x}^{2}}}{{{(1-{{x}^{2}})}^{\dfrac{3}{2}}}}\].
Hence, the second derivative of \[y=x\arcsin x\] is \[\dfrac{2-{{x}^{2}}}{{{(1-{{x}^{2}})}^{\dfrac{3}{2}}}}\].
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 signifies the slope of the function.
Complete step-by-step solution -
To find the second derivative of the function \[y=x\arcsin x\], 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 \[\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left( g(x)\times h(x) \right)=g(x)\times \dfrac{d}{dx}h(x)+h(x)\dfrac{d}{dx}g(x)\].
Substituting \[g(x)=x,h(x)=\arcsin x\] in the above equation, we have \[\dfrac{df(x)}{dx}=\dfrac{d}{dx}(x\arcsin x)=x\times \dfrac{d}{dx}(\arcsin x)+\arcsin x\times \dfrac{d}{dx}(x).....\left( 1 \right)\] .
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 \[\dfrac{d}{dx}g(x)=\dfrac{d}{dx}(x)=1.....\left( 2 \right)\].
We know that differentiation of any function of the form \[y=\arcsin x\] is \[\dfrac{d}{dx}h(x)=\dfrac{d}{dx}(\arcsin x)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}.....\left( 3 \right)\].
Substituting the value of equation (2) and (3) in equation (1), we get \[\dfrac{df(x)}{dx}=x\times \dfrac{d}{dx}(\arcsin x)+\arcsin x\times \dfrac{d}{dx}(x)=x\times \dfrac{1}{\sqrt{1-{{x}^{2}}}}+\arcsin x=\dfrac{x}{\sqrt{1-{{x}^{2}}}}+\arcsin x.....\left( 4 \right)\].
Thus, we have \[\dfrac{dy}{dx}=\dfrac{x}{\sqrt{1-{{x}^{2}}}}+\arcsin x\] 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 \[a(x)=\dfrac{dy}{dx}=\dfrac{x}{\sqrt{1-{{x}^{2}}}}+\arcsin x\].
So, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)\].
We can write \[a(x)\] as a sum of two functions \[a(x)=u(x)+v(x)\].
We know that the sum rule for differentiation of two functions is such that if \[y=a(x)=u(x)+v(x)\] then \[\dfrac{dy}{dx}=\dfrac{d}{dx}a(x)=\dfrac{d}{dx}u(x)+\dfrac{d}{dx}v(x)\].
Substituting \[u(x)=\dfrac{x}{\sqrt{1-{{x}^{2}}}},v(x)=\arcsin x\] in the above equation, we have \[\dfrac{dy}{dx}=\dfrac{d}{dx}a(x)=\dfrac{d}{dx}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)+\dfrac{d}{dx}(\arcsin x).....\left( 5 \right)\].
To find the value of \[\dfrac{d}{dx}u(x)\], 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 \[\alpha (x)=x,\beta (x)=\sqrt{1-{{x}^{2}}}\] in the above equation, we get \[\dfrac{d}{dx}u(x)=\dfrac{\sqrt{1-{{x}^{2}}}\times \dfrac{d}{dx}(x)-x\times \dfrac{d}{dx}(\sqrt{1-{{x}^{2}}})}{{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}.....\left( 6 \right)\].
Let’s assume \[t(x)=\sqrt{1-{{x}^{2}}}\].
We can write \[t(x)\] as a composition of two functions such that \[t(x)=w(z(x))\] where \[w(x)=\sqrt{x,}z(x)=1-{{x}^{2}}\].
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 \[w(x)\] and \[z(x)\] such that \[y=w(z(x))\], then \[\dfrac{dy}{dx}=\dfrac{dw(z(x))}{dz(x)}\times \dfrac{dz(x)}{dx}\].
Substituting \[w(x)=\sqrt{x,}z(x)=1-{{x}^{2}}\] in the above equation, we get \[\dfrac{dy}{dx}=\dfrac{d}{dx}t(x)=\dfrac{dw(z(x))}{dz(x)}\times \dfrac{d}{dx}z(x)=\dfrac{d(\sqrt{1-{{x}^{2}}})}{d(1-{{x}^{2}})}\times \dfrac{d}{dx}(1-{{x}^{2}}).....\left( 7 \right)\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=-1,n=2,b=1\] in the above equation, we get \[\dfrac{d}{dx}(1-{{x}^{2}})=-2x.....\left( 8 \right)\].
To find the value of \[\dfrac{d(\sqrt{1-{{x}^{2}}})}{d(1-{{x}^{2}})}\], let’s assume \[\gamma =1-{{x}^{2}}\].
Thus, we have \[\dfrac{d(\sqrt{1-{{x}^{2}}})}{d(1-{{x}^{2}})}=\dfrac{d}{d\gamma }(\sqrt{\gamma })\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=1,n=\dfrac{1}{2},b=0\] in the above equation, we have \[\dfrac{d}{d\gamma }(\sqrt{\gamma })=\dfrac{1}{2\sqrt{\gamma }}\].
So, we have \[\dfrac{d(\sqrt{1-{{x}^{2}}})}{d(1-{{x}^{2}})}=\dfrac{d}{d\gamma }(\sqrt{\gamma })=\dfrac{1}{2\sqrt{\gamma }}=\dfrac{1}{2\sqrt{1-{{x}^{2}}}}.....\left( 9 \right)\].
Substituting equation (8) and (9) in equation (7), we get \[\dfrac{dy}{dx}=\dfrac{d}{dx}t(x)=\dfrac{d(\sqrt{1-{{x}^{2}}})}{d(1-{{x}^{2}})}\times \dfrac{d}{dx}(1-{{x}^{2}})=\dfrac{1}{2\sqrt{1-{{x}^{2}}}}\times (-2x)=\dfrac{-x}{\sqrt{1-{{x}^{2}}}}.....\left( 10 \right)\].
Substituting equation (2) and (10) in equation (6), we get \[\dfrac{d}{dx}u(x)=\dfrac{\sqrt{1-{{x}^{2}}}\times \dfrac{d}{dx}(x)-x\times \dfrac{d}{dx}(\sqrt{1-{{x}^{2}}})}{{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}=\dfrac{\sqrt{1-{{x}^{2}}}\times 1-x\times \left( \dfrac{-x}{\sqrt{1-{{x}^{2}}}} \right)}{1-{{x}^{2}}}\].
Solving the above equation by taking LCM, we have \[\dfrac{d}{dx}u(x)=\dfrac{\sqrt{1-{{x}^{2}}}\times 1-x\times \left( \dfrac{-x}{\sqrt{1-{{x}^{2}}}} \right)}{1-{{x}^{2}}}=\dfrac{(1-{{x}^{2}})+{{x}^{2}}}{{{(1-{{x}^{2}})}^{\dfrac{3}{2}}}}=\dfrac{1}{{{(1-{{x}^{2}})}^{\dfrac{3}{2}}}}={{(1-{{x}^{2}})}^{\dfrac{-3}{2}}}.....\left( 11 \right)\].
Substituting equation (3) and (11) in equation (5) and solving by taking LCM, we have \[\dfrac{d}{dx}a(x)=\dfrac{d}{dx}\left( \dfrac{x}{\sqrt{1-{{x}^{2}}}} \right)+\dfrac{d}{dx}(\arcsin x)={{(1-{{x}^{2}})}^{\dfrac{-3}{2}}}+\dfrac{1}{\sqrt{1-{{x}^{2}}}}=\dfrac{1+(1-{{x}^{2}})}{{{(1-{{x}^{2}})}^{\dfrac{3}{2}}}}=\dfrac{2-{{x}^{2}}}{{{(1-{{x}^{2}})}^{\dfrac{3}{2}}}}\].
Hence, the second derivative of \[y=x\arcsin x\] is \[\dfrac{2-{{x}^{2}}}{{{(1-{{x}^{2}})}^{\dfrac{3}{2}}}}\].
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 signifies the slope of the function.
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