Question

Find the second derivative of the following function. $y={{e}^{\sqrt{x}}}\sqrt{{{x}^{2}}-1}$

Answer 1

- Hint: Use product rule and chain rule of composition of two functions to differentiate the given function twice. Use the fact that differentiation of function $y={{e}^{x}}$ is $\dfrac{dy}{dx}={{e}^{x}}$ and $y=a{{x}^{n}}+b$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.

Complete step-by-step solution -

To find the second derivative of the function$y={{e}^{\sqrt{x}}}\sqrt{{{x}^{2}}-1}$, 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 the product rule of differentiation of product of two functions which states that$\dfrac{d}{dx}a(x)=\dfrac{d}{dx}\left( u(x)\times v(x) \right)=v(x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}v(x)$.
Substituting$u(x)={{e}^{\sqrt{x}}},v(x)=\sqrt{{{x}^{2}}-1}$in the above equation, we have$\dfrac{d}{dx}a(x)=\dfrac{d}{dx}\left( {{e}^{\sqrt{x}}}\sqrt{{{x}^{2}}-1} \right)=\sqrt{{{x}^{2}}-1}\times \dfrac{d}{dx}({{e}^{\sqrt{x}}})+{{e}^{\sqrt{x}}}\dfrac{d}{dx}(\sqrt{{{x}^{2}}-1})$.$...(1)$
To find the value of$\dfrac{d}{dx}({{e}^{\sqrt{x}}})$, let’s assume$u(x)={{e}^{\sqrt{x}}}$as a composition of two functions such that$u(x)=f(g(x))$where$f(x)={{e}^{x}},g(x)=\sqrt{x}$.
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$f(x)$and$g(x)$such that$y=f(g(x))$, then$\dfrac{dy}{dx}=\dfrac{df(g(x))}{dg(x)}\times \dfrac{dg(x)}{dx}$.
Substituting$f(x)={{e}^{x}},g(x)=\sqrt{x}$in the above equation, we get$\dfrac{d}{dx}u(x)=\dfrac{df(g(x))}{dg(x)}\times \dfrac{dg(x)}{dx}=\dfrac{d{{e}^{\sqrt{x}}}}{d(\sqrt{x})}\times \dfrac{d(\sqrt{x})}{dx}$. $...(2)$
To find the value of$\dfrac{d{{e}^{\sqrt{x}}}}{d(\sqrt{x})}$, let’s assume$t=\sqrt{x}$.
Thus, we have$\dfrac{d{{e}^{\sqrt{x}}}}{d(\sqrt{x})}=\dfrac{d({{e}^{t}})}{dt}$.
We know that differentiation of any function of the form$y={{e}^{x}}$is$\dfrac{dy}{dx}={{e}^{x}}$.
So, we have$\dfrac{d{{e}^{\sqrt{x}}}}{d(\sqrt{x})}=\dfrac{d({{e}^{t}})}{dt}={{e}^{t}}={{e}^{\sqrt{x}}}$. $...(3)$
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$\dfrac{d(\sqrt{x})}{dx}=\dfrac{1}{2\sqrt{x}}$. $...(4)$
Substituting equation$(3)$and$(4)$in equation$(2)$, we get$\dfrac{d}{dx}u(x)=\dfrac{d{{e}^{\sqrt{x}}}}{d(\sqrt{x})}\times \dfrac{d(\sqrt{x})}{dx}={{e}^{\sqrt{x}}}\times \dfrac{1}{2\sqrt{x}}=\dfrac{{{e}^{\sqrt{x}}}}{2\sqrt{x}}$. $...(5)$
To find the value of$\dfrac{d}{dx}(\sqrt{{{x}^{2}}-1})$, let’s assume$v(x)=\sqrt{{{x}^{2}}-1}$ as a composition of two functions such that$v(x)=\alpha (\beta (x))$ where$\alpha (x)=\sqrt{x},\beta (x)={{x}^{2}}-1$.
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$\alpha (x)$and$\beta (x)$such that$y=\alpha (\beta (x))$, then$\dfrac{dy}{dx}=\dfrac{d\alpha (\beta (x))}{d\beta (x)}\times \dfrac{d\beta (x)}{dx}$.
Substituting$\alpha (x)=\sqrt{x},\beta (x)={{x}^{2}}-1$in the above equation, we have$\dfrac{dy}{dx}=\dfrac{d\alpha (\beta (x))}{d\beta (x)}\times \dfrac{d\beta (x)}{dx}=\dfrac{d\left( \sqrt{{{x}^{2}}-1} \right)}{d({{x}^{2}}-1)}\times \dfrac{d({{x}^{2}}-1)}{dx}$. $...\left( 6 \right)$
To find the value of$\dfrac{d\left( \sqrt{{{x}^{2}}-1} \right)}{d({{x}^{2}}-1)}$, let’s assume$z={{x}^{2}}-1$.
Thus, we have$\dfrac{d\left( \sqrt{{{x}^{2}}-1} \right)}{d({{x}^{2}}-1)}=\dfrac{d(\sqrt{z)}}{dz}$.
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$\dfrac{d(\sqrt{x})}{dx}=\dfrac{1}{2\sqrt{x}}$.
So, we have$\dfrac{d\left( \sqrt{{{x}^{2}}-1} \right)}{d({{x}^{2}}-1)}=\dfrac{d(\sqrt{z)}}{dz}=\dfrac{1}{2\sqrt{z}}=\dfrac{1}{2\sqrt{{{x}^{2}}-1}}$. $...\left( 7 \right)$
To find the value of$\dfrac{d({{x}^{2}}-1)}{dx}$, substitute$a=1,n=2,b=-1$in the differentiation of any function of the form$y=a{{x}^{n}}+b$is$\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Thus, we have$\dfrac{d({{x}^{2}}-1)}{dx}=2x$. $...\left( 8 \right)$
Substituting equation$\left( 7 \right)$and$\left( 8 \right)$in equation$(6)$, we get$\dfrac{dy}{dx}=\dfrac{d\left( \sqrt{{{x}^{2}}-1} \right)}{d({{x}^{2}}-1)}\times \dfrac{d({{x}^{2}}-1)}{dx}=\dfrac{1}{2\sqrt{{{x}^{2}}-1}}\times 2x=\dfrac{x}{\sqrt{{{x}^{2}}-1}}$. $...(9)$
Substituting equation$(5)$and$(9)$in equation$(1)$, we get$\dfrac{d}{dx}a(x)=\dfrac{d}{dx}\left( {{e}^{\sqrt{x}}}\sqrt{{{x}^{2}}-1} \right)=\sqrt{{{x}^{2}}-1}\times \dfrac{d}{dx}({{e}^{\sqrt{x}}})+{{e}^{\sqrt{x}}}\dfrac{d}{dx}(\sqrt{{{x}^{2}}-1})=\sqrt{{{x}^{2}}-1}\times \dfrac{{{e}^{\sqrt{x}}}}{2\sqrt{x}}+{{e}^{\sqrt{x}}}\times \dfrac{x}{\sqrt{{{x}^{2}}-1}}$
Solving the above equation, we get$\dfrac{d}{dx}a(x)={{e}^{\sqrt{x}}}\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)$. $...(10)$
Thus, we have$\dfrac{dy}{dx}={{e}^{\sqrt{x}}}\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)$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$h(x)=\dfrac{dy}{dx}={{e}^{\sqrt{x}}}\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)$.
So, we have$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}h(x)$.
We can write$h(x)$as a product of two functions of the form$h(x)=u(x)\times \gamma (x)$where$u(x)={{e}^{\sqrt{x}}}$and$\gamma (x)=\dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}}$.
We will use the product rule of differentiation of product of two functions which states that$\dfrac{d}{dx}h(x)=\dfrac{d}{dx}\left( u(x)\times \gamma (x) \right)=\gamma (x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}\gamma (x)$.
Substituting$u(x)={{e}^{\sqrt{x}}},\gamma (x)=\dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}}$in the above equation, we have$\dfrac{d}{dx}h(x)=\gamma (x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}\gamma (x)=\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)\times \dfrac{d({{e}^{\sqrt{x}}})}{dx}+{{e}^{\sqrt{x}}}\times \dfrac{d}{dx}\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)$
$...(11)$
To find the value of$\dfrac{d}{dx}\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)$, we can write$\gamma (x)=\dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}}$as sum of two functions${{\alpha }_{1}}(x)$and${{\alpha }_{2}}(x)$.
We can use sum rule of differentiation of two functions which states that if$y={{\alpha }_{1}}(x)+{{\alpha }_{2}}(x)$, then$\dfrac{dy}{dx}=\dfrac{d}{dx}{{\alpha }_{1}}(x)+\dfrac{d}{dx}{{\alpha }_{2}}(x)$.
So, we have$\dfrac{d}{dx}\gamma (x)=\dfrac{d}{dx}{{\alpha }_{1}}(x)+\dfrac{d}{dx}{{\alpha }_{2}}(x)=\dfrac{d}{dx}\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}} \right)+\dfrac{d}{dx}\left( \dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)$ $...\left( 12 \right)$
To find the value of$\dfrac{d}{dx}\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}} \right)$, we will assume it as a quotient of two functions$\dfrac{v(x)}{{{\beta }_{1}}(x)}$where$v(x)=\sqrt{{{x}^{2}}-1},{{\beta }_{1}}(x)=2\sqrt{x}$.
We will use quotient rule of differentiation which states that if$y=\dfrac{v(x)}{{{\beta }_{1}}(x)}$then, we have$\dfrac{dy}{dx}=\dfrac{{{\beta }_{1}}(x)\dfrac{d}{dx}v(x)-v(x)\dfrac{d}{dx}{{\beta }_{1}}(x)}{\beta _{1}^{2}(x)}$.
Substituting$v(x)=\sqrt{{{x}^{2}}-1},{{\beta }_{1}}(x)=2\sqrt{x}$in the above equation, we have$\dfrac{d}{dx}{{\alpha }_{1}}\left( x \right)=\dfrac{{{\beta }_{1}}(x)\dfrac{d}{dx}v(x)-v(x)\dfrac{d}{dx}{{\beta }_{1}}(x)}{\beta _{1}^{2}(x)}=\dfrac{2\sqrt{x}\times \dfrac{d}{dx}\left( \sqrt{{{x}^{2}}-1} \right)-\left( \sqrt{{{x}^{2}}-1} \right)\times \dfrac{d}{dx}\left( 2\sqrt{x} \right)}{{{\left( 2\sqrt{x} \right)}^{2}}}$ $...(13)$
To find the value of$\dfrac{d}{dx}{{\beta }_{1}}(x)$, substitute$a=2,n=\dfrac{1}{2},b=0$in the differentiation of any function of the form$y=a{{x}^{n}}+b$is$\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Thus, we have$\dfrac{d}{dx}{{\beta }_{1}}(x)=\dfrac{d}{dx}(2\sqrt{x})=\dfrac{1}{\sqrt{x}}$. $...(14)$
Substituting equation$\left( 9 \right)$and$(14)$in equation$\left( 13 \right)$, we have$\dfrac{d}{dx}{{\alpha }_{1}}\left( x \right)=\dfrac{2\sqrt{x}\times \dfrac{d}{dx}\left( \sqrt{{{x}^{2}}-1} \right)-\left( \sqrt{{{x}^{2}}-1} \right)\times \dfrac{d}{dx}\left( 2\sqrt{x} \right)}{{{\left( 2\sqrt{x} \right)}^{2}}}=\dfrac{2\sqrt{x}\times \dfrac{x}{\sqrt{{{x}^{2}}-1}}-\dfrac{1}{\sqrt{x}}\times \sqrt{{{x}^{2}}-1}}{4x}$.
Solving the above equation by taking LCM, we get$\dfrac{d}{dx}{{\alpha }_{1}}\left( x \right)=\dfrac{{{x}^{2}}+1}{4{{x}^{\dfrac{3}{2}}}\sqrt{{{x}^{2}}-1}}$. $...\left( 15 \right)$
To find the value of$\dfrac{d}{dx}\left( \dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)$, we will assume it as a quotient of two functions$\dfrac{{{\beta }_{2}}(x)}{v(x)}$where$v(x)=\sqrt{{{x}^{2}}-1},{{\beta }_{2}}(x)=x$.
We will use quotient rule of differentiation which states that if$y=\dfrac{{{\beta }_{2}}(x)}{v(x)}$then, we have$\dfrac{dy}{dx}=\dfrac{v(x)\dfrac{d}{dx}{{\beta }_{2}}(x)-{{\beta }_{2}}(x)\dfrac{d}{dx}v(x)}{{{v}^{2}}(x)}$.
Substituting$v(x)=\sqrt{{{x}^{2}}-1},{{\beta }_{2}}(x)=x$in the above equation, we get$\dfrac{dy}{dx}=\dfrac{v(x)\dfrac{d}{dx}{{\beta }_{2}}(x)-{{\beta }_{2}}(x)\dfrac{d}{dx}v(x)}{{{v}^{2}}(x)}=\dfrac{\sqrt{{{x}^{2}}-1}\times \dfrac{d}{dx}\left( x \right)-x\times \dfrac{d}{dx}\left( \sqrt{{{x}^{2}}-1} \right)}{{{\left( \sqrt{{{x}^{2}}-1} \right)}^{2}}}$. $...\left( 16 \right)$
To find the value of$\dfrac{d}{dx}{{\beta }_{2}}(x)$, substitute$a=1,n=1,b=0$in the differentiation of any function of the form$y=a{{x}^{n}}+b$is$\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Thus, we have$\dfrac{d}{dx}{{\beta }_{2}}\left( x \right)=\dfrac{d}{dx}\left( x \right)=1$. $...\left( 17 \right)$
Substituting equation$\left( 9 \right)$and$\left( 17 \right)$in equation$\left( 16 \right)$, we have$\dfrac{d}{dx}{{\alpha }_{2}}(x)=\dfrac{\sqrt{{{x}^{2}}-1}\times \dfrac{d}{dx}\left( x \right)-x\times \dfrac{d}{dx}\left( \sqrt{{{x}^{2}}-1} \right)}{{{\left( \sqrt{{{x}^{2}}-1} \right)}^{2}}}=\dfrac{\sqrt{{{x}^{2}}-1}\times 1-x\times \dfrac{x}{\sqrt{{{x}^{2}}-1}}}{{{x}^{2}}-1}$.
Solving the above equation by taking LCM, we get$\dfrac{d}{dx}{{\alpha }_{2}}\left( x \right)=\dfrac{-1}{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}}$. $...\left( 18 \right)$
Substituting equation$\left( 15 \right)$and$\left( 18 \right)$in equation$\left( 12 \right)$, we get$\dfrac{d}{dx}\gamma (x)=\dfrac{d}{dx}\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}} \right)+\dfrac{d}{dx}\left( \dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)=\dfrac{{{x}^{2}}+1}{4{{x}^{\dfrac{3}{2}}}\sqrt{{{x}^{2}}-1}}+\dfrac{-1}{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}}$. $...\left( 19 \right)$
Substituting equation$\left( 5 \right)$and$\left( 19 \right)$in equation$\left( 11 \right)$, we get$\dfrac{d}{dx}h(x)=\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)\times \dfrac{d({{e}^{\sqrt{x}}})}{dx}+{{e}^{\sqrt{x}}}\times \dfrac{d}{dx}\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)=\left( \dfrac{\sqrt{{{x}^{2}}-1}}{2\sqrt{x}}+\dfrac{x}{\sqrt{{{x}^{2}}-1}} \right)\times \dfrac{{{e}^{\sqrt{x}}}}{2\sqrt{x}}+{{e}^{\sqrt{x}}}\times \left( \dfrac{{{x}^{2}}+1}{4{{x}^{\dfrac{3}{2}}}\sqrt{{{x}^{2}}-1}}+\dfrac{-1}{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}} \right)$
Solving the above equation, we get$\dfrac{d}{dx}h(x)=\left( \dfrac{\sqrt{{{x}^{2}}-1}}{4x}+\dfrac{\sqrt{x}}{2\sqrt{{{x}^{2}}-1}} \right)\times {{e}^{\sqrt{x}}}+{{e}^{\sqrt{x}}}\times \left( \dfrac{{{x}^{2}}+1}{4{{x}^{\dfrac{3}{2}}}\sqrt{{{x}^{2}}-1}}+\dfrac{-1}{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}} \right)$.
Further simplifying the equation, we get$\dfrac{d}{dx}h(x)=\left( \dfrac{\sqrt{{{x}^{2}}-1}}{4x}+\dfrac{\sqrt{x}}{2\sqrt{{{x}^{2}}-1}}+\dfrac{{{x}^{2}}+1}{4{{x}^{\dfrac{3}{2}}}\sqrt{{{x}^{2}}-1}}+\dfrac{-1}{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}} \right)\times {{e}^{\sqrt{x}}}$.
Thus, we have$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}h(x)=\left( \dfrac{\sqrt{{{x}^{2}}-1}}{4x}+\dfrac{\sqrt{x}}{2\sqrt{{{x}^{2}}-1}}+\dfrac{{{x}^{2}}+1}{4{{x}^{\dfrac{3}{2}}}\sqrt{{{x}^{2}}-1}}+\dfrac{-1}{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}} \right)\times {{e}^{\sqrt{x}}}$.
Hence, the second derivative of $y={{e}^{\sqrt{x}}}\sqrt{{{x}^{2}}-1}$is$\left( \dfrac{\sqrt{{{x}^{2}}-1}}{4x}+\dfrac{\sqrt{x}}{2\sqrt{{{x}^{2}}-1}}+\dfrac{{{x}^{2}}+1}{4{{x}^{\dfrac{3}{2}}}\sqrt{{{x}^{2}}-1}}+\dfrac{-1}{{{\left( {{x}^{2}}-1 \right)}^{\dfrac{3}{2}}}} \right)\times {{e}^{\sqrt{x}}}$.
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.