Find the second derivative of the following function. \[y=\arctan \sqrt{{{e}^{x}}-1}\]
Answer
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Hint: Use chain rule of composition of three functions to differentiate the given function twice. Use the fact that differentiation of function of the form \[y=a{{e}^{x}}+b\] is \[\dfrac{dy}{dx}=a{{e}^{x}}\],\[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\] and \[y=\arctan x\] is \[\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\].
Complete step-by-step answer:
To find the second derivative of the function \[y=\arctan \sqrt{{{e}^{x}}-1}\], 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 chain rule of differentiation of composition of three functions to find the derivative which states that if y is a composition of three functions \[f(x)\],\[g(x)\] and \[h(x)\] such that \[y=f(g(h(x)))\], then \[\dfrac{dy}{dx}=\dfrac{df(g(h(x)))}{dg(h(x))}\times \dfrac{dg(h(x))}{dh(x)}\times \dfrac{dh(x)}{dx}\].
Substituting \[f(x)=\arctan x\], \[g(x)=\sqrt{x}\] and \[h(x)={{e}^{x}}-1\] in the above equation, we get \[\dfrac{dy}{dx}=\dfrac{d\arctan \sqrt{{{e}^{x}}-1}}{d\sqrt{{{e}^{x}}-1}}\times \dfrac{d\sqrt{{{e}^{x}}-1}}{d({{e}^{x}}-1)}\times \dfrac{d}{dx}({{e}^{x}}-1).....\left( 1 \right)\].
We know that differentiation of any function of the form \[y=a{{e}^{x}}+b\] is \[\dfrac{dy}{dx}=a{{e}^{x}}\].
Substituting \[a=1,b=0\] in the above equation, we get \[\dfrac{d}{dx}h(x)=\dfrac{d}{dx}({{e}^{x}}-1)={{e}^{x}}.....\left( 2 \right)\].
To find the value of \[\dfrac{d\sqrt{{{e}^{x}}-1}}{d({{e}^{x}}-1)}\], we will substitute \[t={{e}^{x}}-1\].
Thus, we have \[\dfrac{d\sqrt{{{e}^{x}}-1}}{d({{e}^{x}}-1)}=\dfrac{d\sqrt{t}}{dt}\].
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 get \[\dfrac{d\sqrt{t}}{dt}=\dfrac{1}{2\sqrt{t}}\].
Thus, we have \[\dfrac{d\sqrt{{{e}^{x}}-1}}{d({{e}^{x}}-1)}=\dfrac{d\sqrt{t}}{dt}=\dfrac{1}{2\sqrt{t}}=\dfrac{1}{2\sqrt{{{e}^{x}}-1}}.....\left( 3 \right)\].
To find the value of \[\dfrac{d\arctan \sqrt{{{e}^{x}}-1}}{d\sqrt{{{e}^{x}}-1}}\], let’s assume \[z=\sqrt{{{e}^{x}}-1}\].
Thus, we have \[\dfrac{d\arctan \sqrt{{{e}^{x}}-1}}{d\sqrt{{{e}^{x}}-1}}=\dfrac{d\arctan z}{dz}\].
We know that differentiation of any function of the form \[y=\arctan x\] is \[\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\].
So, we have \[\dfrac{d\arctan \sqrt{{{e}^{x}}-1}}{d\sqrt{{{e}^{x}}-1}}=\dfrac{d\arctan z}{dz}=\dfrac{1}{1+{{z}^{2}}}=\dfrac{1}{1+{{e}^{x}}-1}=\dfrac{1}{{{e}^{x}}}....\left( 4 \right)\].
Substituting equation \[(2)\], \[(3)\] and \[(4)\] in equation \[(1)\], we get \[\dfrac{dy}{dx}=\dfrac{d\arctan \sqrt{{{e}^{x}}-1}}{d\sqrt{{{e}^{x}}-1}}\times \dfrac{d\sqrt{{{e}^{x}}-1}}{d({{e}^{x}}-1)}\times \dfrac{d}{dx}({{e}^{x}}-1)=\dfrac{1}{{{e}^{x}}}\times \dfrac{1}{2\sqrt{{{e}^{x}}-1}}\times {{e}^{x}}=\dfrac{1}{2\sqrt{{{e}^{x}}-1}}\].
Thus, we have \[\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{{{e}^{x}}-1}}\] 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 \[a(x)=\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{{{e}^{x}}-1}}=\dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}}\].
So, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)\].
We can rewrite \[a(x)=v\left( u\left( x \right) \right)\] as a composition of two functions \[u(x)\] and \[v(x)\] such that \[u(x)={{e}^{x}}-1\] and \[v(x)=\dfrac{1}{2}{{x}^{\dfrac{-1}{2}}}\].
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 \[u(x)\] and \[v(x)\] such that \[y=v\left( u\left( x \right) \right)\], then \[\dfrac{dy}{dx}=\dfrac{dv\left( u\left( x \right) \right)}{du\left( x \right)}\times \dfrac{du(x)}{dx}\].
Substituting \[u(x)={{e}^{x}}-1\] and \[v(x)=\dfrac{1}{2}{{x}^{\dfrac{-1}{2}}}\] in the above equation, we get \[\dfrac{da(x)}{dx}=\dfrac{dy}{dx}=\dfrac{d\left( \dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}} \right)}{d\left( {{e}^{x}}-1 \right)}\times \dfrac{d\left( {{e}^{x}}-1 \right)}{dx}.....\left( 5 \right)\].
We know that differentiation of any function of the form \[y=a{{e}^{x}}+b\] is \[\dfrac{dy}{dx}=a{{e}^{x}}\].
Substituting \[a=1,b=0\] in the above equation, we get \[\dfrac{d}{dx}u(x)=\dfrac{d}{dx}({{e}^{x}}-1)={{e}^{x}}.....\left( 6 \right)\].
To find the value of \[\dfrac{d\left( \dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}} \right)}{d\left( {{e}^{x}}-1 \right)}\], we will assume $t={{e}^{x}}-1$.
We can rewrite the above equation as \[\dfrac{d\left( \dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}} \right)}{d\left( {{e}^{x}}-1 \right)}=\dfrac{d\left( \dfrac{1}{2}{{t}^{\dfrac{-1}{2}}} \right)}{dt}\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=\dfrac{1}{2},n=\dfrac{-1}{2},b=0\] in the above equation, we have \[\dfrac{d\left( \dfrac{1}{2}{{t}^{\dfrac{-1}{2}}} \right)}{dt}=\dfrac{1}{2}\left( \dfrac{-1}{2} \right){{t}^{\dfrac{-3}{2}}}=\dfrac{-1}{4}{{t}^{\dfrac{-3}{2}}}\].
So, we have \[\dfrac{d\left( \dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}} \right)}{d\left( {{e}^{x}}-1 \right)}=\dfrac{d\left( \dfrac{1}{2}{{t}^{\dfrac{-1}{2}}} \right)}{dt}=\dfrac{-1}{4}{{t}^{\dfrac{-3}{2}}}=\dfrac{-1}{4}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-3}{2}}}.....\left( 7 \right)\].
Substituting the value of equation \[(6)\] and \[(7)\] in equation \[(5)\], we have \[\dfrac{da(x)}{dx}=\dfrac{dy}{dx}=\dfrac{d\left( \dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}} \right)}{d\left( {{e}^{x}}-1 \right)}\times \dfrac{d\left( {{e}^{x}}-1 \right)}{dx}=\dfrac{-1}{4}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-3}{2}}}\times {{e}^{x}}=\dfrac{-{{e}^{x}}}{4{{\left( {{e}^{x}}-1 \right)}^{\dfrac{3}{2}}}}\].
Thus, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)=\dfrac{-{{e}^{x}}}{4{{\left( {{e}^{x}}-1 \right)}^{\dfrac{3}{2}}}}\].
Hence, the second derivative of \[y=\arctan \sqrt{{{e}^{x}}-1}\] is \[\dfrac{-{{e}^{x}}}{4{{\left( {{e}^{x}}-1 \right)}^{\dfrac{3}{2}}}}\].
Note: We can also solve this question using First Principle to find the derivative. One must know that \[\arctan x={{\tan }^{-1}}x\]. 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 answer:
To find the second derivative of the function \[y=\arctan \sqrt{{{e}^{x}}-1}\], 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 chain rule of differentiation of composition of three functions to find the derivative which states that if y is a composition of three functions \[f(x)\],\[g(x)\] and \[h(x)\] such that \[y=f(g(h(x)))\], then \[\dfrac{dy}{dx}=\dfrac{df(g(h(x)))}{dg(h(x))}\times \dfrac{dg(h(x))}{dh(x)}\times \dfrac{dh(x)}{dx}\].
Substituting \[f(x)=\arctan x\], \[g(x)=\sqrt{x}\] and \[h(x)={{e}^{x}}-1\] in the above equation, we get \[\dfrac{dy}{dx}=\dfrac{d\arctan \sqrt{{{e}^{x}}-1}}{d\sqrt{{{e}^{x}}-1}}\times \dfrac{d\sqrt{{{e}^{x}}-1}}{d({{e}^{x}}-1)}\times \dfrac{d}{dx}({{e}^{x}}-1).....\left( 1 \right)\].
We know that differentiation of any function of the form \[y=a{{e}^{x}}+b\] is \[\dfrac{dy}{dx}=a{{e}^{x}}\].
Substituting \[a=1,b=0\] in the above equation, we get \[\dfrac{d}{dx}h(x)=\dfrac{d}{dx}({{e}^{x}}-1)={{e}^{x}}.....\left( 2 \right)\].
To find the value of \[\dfrac{d\sqrt{{{e}^{x}}-1}}{d({{e}^{x}}-1)}\], we will substitute \[t={{e}^{x}}-1\].
Thus, we have \[\dfrac{d\sqrt{{{e}^{x}}-1}}{d({{e}^{x}}-1)}=\dfrac{d\sqrt{t}}{dt}\].
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 get \[\dfrac{d\sqrt{t}}{dt}=\dfrac{1}{2\sqrt{t}}\].
Thus, we have \[\dfrac{d\sqrt{{{e}^{x}}-1}}{d({{e}^{x}}-1)}=\dfrac{d\sqrt{t}}{dt}=\dfrac{1}{2\sqrt{t}}=\dfrac{1}{2\sqrt{{{e}^{x}}-1}}.....\left( 3 \right)\].
To find the value of \[\dfrac{d\arctan \sqrt{{{e}^{x}}-1}}{d\sqrt{{{e}^{x}}-1}}\], let’s assume \[z=\sqrt{{{e}^{x}}-1}\].
Thus, we have \[\dfrac{d\arctan \sqrt{{{e}^{x}}-1}}{d\sqrt{{{e}^{x}}-1}}=\dfrac{d\arctan z}{dz}\].
We know that differentiation of any function of the form \[y=\arctan x\] is \[\dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}\].
So, we have \[\dfrac{d\arctan \sqrt{{{e}^{x}}-1}}{d\sqrt{{{e}^{x}}-1}}=\dfrac{d\arctan z}{dz}=\dfrac{1}{1+{{z}^{2}}}=\dfrac{1}{1+{{e}^{x}}-1}=\dfrac{1}{{{e}^{x}}}....\left( 4 \right)\].
Substituting equation \[(2)\], \[(3)\] and \[(4)\] in equation \[(1)\], we get \[\dfrac{dy}{dx}=\dfrac{d\arctan \sqrt{{{e}^{x}}-1}}{d\sqrt{{{e}^{x}}-1}}\times \dfrac{d\sqrt{{{e}^{x}}-1}}{d({{e}^{x}}-1)}\times \dfrac{d}{dx}({{e}^{x}}-1)=\dfrac{1}{{{e}^{x}}}\times \dfrac{1}{2\sqrt{{{e}^{x}}-1}}\times {{e}^{x}}=\dfrac{1}{2\sqrt{{{e}^{x}}-1}}\].
Thus, we have \[\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{{{e}^{x}}-1}}\] 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 \[a(x)=\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{{{e}^{x}}-1}}=\dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}}\].
So, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)\].
We can rewrite \[a(x)=v\left( u\left( x \right) \right)\] as a composition of two functions \[u(x)\] and \[v(x)\] such that \[u(x)={{e}^{x}}-1\] and \[v(x)=\dfrac{1}{2}{{x}^{\dfrac{-1}{2}}}\].
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 \[u(x)\] and \[v(x)\] such that \[y=v\left( u\left( x \right) \right)\], then \[\dfrac{dy}{dx}=\dfrac{dv\left( u\left( x \right) \right)}{du\left( x \right)}\times \dfrac{du(x)}{dx}\].
Substituting \[u(x)={{e}^{x}}-1\] and \[v(x)=\dfrac{1}{2}{{x}^{\dfrac{-1}{2}}}\] in the above equation, we get \[\dfrac{da(x)}{dx}=\dfrac{dy}{dx}=\dfrac{d\left( \dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}} \right)}{d\left( {{e}^{x}}-1 \right)}\times \dfrac{d\left( {{e}^{x}}-1 \right)}{dx}.....\left( 5 \right)\].
We know that differentiation of any function of the form \[y=a{{e}^{x}}+b\] is \[\dfrac{dy}{dx}=a{{e}^{x}}\].
Substituting \[a=1,b=0\] in the above equation, we get \[\dfrac{d}{dx}u(x)=\dfrac{d}{dx}({{e}^{x}}-1)={{e}^{x}}.....\left( 6 \right)\].
To find the value of \[\dfrac{d\left( \dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}} \right)}{d\left( {{e}^{x}}-1 \right)}\], we will assume $t={{e}^{x}}-1$.
We can rewrite the above equation as \[\dfrac{d\left( \dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}} \right)}{d\left( {{e}^{x}}-1 \right)}=\dfrac{d\left( \dfrac{1}{2}{{t}^{\dfrac{-1}{2}}} \right)}{dt}\].
We know that differentiation of any function of the form \[y=a{{x}^{n}}+b\] is \[\dfrac{dy}{dx}=an{{x}^{n-1}}\].
Substituting \[a=\dfrac{1}{2},n=\dfrac{-1}{2},b=0\] in the above equation, we have \[\dfrac{d\left( \dfrac{1}{2}{{t}^{\dfrac{-1}{2}}} \right)}{dt}=\dfrac{1}{2}\left( \dfrac{-1}{2} \right){{t}^{\dfrac{-3}{2}}}=\dfrac{-1}{4}{{t}^{\dfrac{-3}{2}}}\].
So, we have \[\dfrac{d\left( \dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}} \right)}{d\left( {{e}^{x}}-1 \right)}=\dfrac{d\left( \dfrac{1}{2}{{t}^{\dfrac{-1}{2}}} \right)}{dt}=\dfrac{-1}{4}{{t}^{\dfrac{-3}{2}}}=\dfrac{-1}{4}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-3}{2}}}.....\left( 7 \right)\].
Substituting the value of equation \[(6)\] and \[(7)\] in equation \[(5)\], we have \[\dfrac{da(x)}{dx}=\dfrac{dy}{dx}=\dfrac{d\left( \dfrac{1}{2}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-1}{2}}} \right)}{d\left( {{e}^{x}}-1 \right)}\times \dfrac{d\left( {{e}^{x}}-1 \right)}{dx}=\dfrac{-1}{4}{{\left( {{e}^{x}}-1 \right)}^{\dfrac{-3}{2}}}\times {{e}^{x}}=\dfrac{-{{e}^{x}}}{4{{\left( {{e}^{x}}-1 \right)}^{\dfrac{3}{2}}}}\].
Thus, we have \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)=\dfrac{-{{e}^{x}}}{4{{\left( {{e}^{x}}-1 \right)}^{\dfrac{3}{2}}}}\].
Hence, the second derivative of \[y=\arctan \sqrt{{{e}^{x}}-1}\] is \[\dfrac{-{{e}^{x}}}{4{{\left( {{e}^{x}}-1 \right)}^{\dfrac{3}{2}}}}\].
Note: We can also solve this question using First Principle to find the derivative. One must know that \[\arctan x={{\tan }^{-1}}x\]. 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.
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