Question

Find the second derivative of the following function. $y={{\ln }^{4}}({{x}^{2}}+1)$.

Answer 1

We have to differentiate the given function twice to get the second derivative. To find the first derivative, we can see that the given function is a composition of functions like $\ln x, ax^n+b$. So, we can use the chain rule and also use the differentiation formulas given by $y=a{{x}^{n}}+b$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$ and $y=\ln x$ is $\dfrac{dy}{dx}=\dfrac{1}{x}$. Now, we will obtain the first derivative as a fraction, so we can apply the quotient rule and then simplify to get the second derivative.

Complete step by step answer:
To find the second derivative of the function $y={{\ln }^{4}}({{x}^{2}}+1)$, 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 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 $f(x)={{x}^{4}}$, $g(x)=\ln x$ and $h(x)={{x}^{2}}+1$ in the above equation, we get $\dfrac{dy}{dx}=\dfrac{d{{\ln }^{4}}({{x}^{2}}+1)}{d\ln ({{x}^{2}}+1)}\times \dfrac{d\ln ({{x}^{2}}+1)}{d({{x}^{2}}+1)}\times \dfrac{d}{dx}({{x}^{2}}+1)....\left( 1 \right)$.
We know that differentiation of any function of the form $y=a{{x}^{n}}+b$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Substituting $a=1,n=2,b=1$ in the above equation, we get $\dfrac{d}{dx}h(x)=\dfrac{d({{x}^{2}}+1)}{dx}=2x.....\left( 2 \right)$.
To find the value of $\dfrac{d\ln ({{x}^{2}}+1)}{d({{x}^{2}}+1)}$, let’s assume $t={{x}^{2}}+1$.
Thus, we have $\dfrac{d\ln ({{x}^{2}}+1)}{d({{x}^{2}}+1)}=\dfrac{d\ln t}{dt}$.
We know that differentiation of function $y=\ln x$ is $\dfrac{dy}{dx}=\dfrac{1}{x}$.
So, we will get $\dfrac{d\ln ({{x}^{2}}+1)}{d({{x}^{2}}+1)}=\dfrac{d\ln t}{dt}=\dfrac{1}{t}=\dfrac{1}{{{x}^{2}}+1}.....\left( 3 \right)$.
To find the value of $\dfrac{d{{\ln }^{4}}({{x}^{2}}+1)}{d\ln ({{x}^{2}}+1)}$, let’s assume $z=\ln ({{x}^{2}}+1)$.
Thus, we have $\dfrac{d{{\ln }^{4}}({{x}^{2}}+1)}{d\ln ({{x}^{2}}+1)}=\dfrac{d({{z}^{4}})}{dz}$.
We know that differentiation of any function of the form $y=a{{x}^{n}}+b$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Substituting $a=1,n=4,b=0$ in the above equation, we get $\dfrac{d{{\ln }^{4}}({{x}^{2}}+1)}{d\ln ({{x}^{2}}+1)}=\dfrac{d({{z}^{4}})}{dz}=4{{z}^{3}}=4{{\ln }^{3}}({{x}^{2}}+1).....\left( 4 \right)$.
Substituting equation $(2)$, $(3)$ and $(4)$ in equation $(1)$, we get $\dfrac{dy}{dx}=\dfrac{d{{\ln }^{4}}({{x}^{2}}+1)}{d\ln ({{x}^{2}}+1)}\times \dfrac{d\ln ({{x}^{2}}+1)}{d({{x}^{2}}+1)}\times \dfrac{d}{dx}({{x}^{2}}+1)=\dfrac{4{{\ln }^{3}}({{x}^{2}}+1)}{{{x}^{2}}+1}\times 2x=\dfrac{8x{{\ln }^{3}}({{x}^{2}}+1)}{{{x}^{2}}+1}$.
Thus, we have $\dfrac{dy}{dx}=\dfrac{8x{{\ln }^{3}}({{x}^{2}}+1)}{{{x}^{2}}+1}$ 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 $a(x)=\dfrac{dy}{dx}=\dfrac{8x{{\ln }^{3}}({{x}^{2}}+1)}{{{x}^{2}}+1}$.
So, we have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)$.
We can rewrite $a(x)$ as a product of two functions $u(x)$ and $v(x)$ such that $u(x)=\dfrac{8x}{{{x}^{2}}+1}$ and $v(x)={{\ln }^{3}}({{x}^{2}}+1)$.
We will use the product rule of differentiation of product of two functions which states that $\dfrac{d}{dx}a(x)=\dfrac{d}{dx}\left( u(x)\times v(x) \right)=v(x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}v(x).....\left( 5 \right)$.
We will now differentiate the functions $u(x)$ and $v(x)$.
We will use chain rule of differentiation of composition of three functions to differentiate the function $v(x)$ which is a composition of three functions $f(x)$, $g(x)$ and $h(x)$ such that $y=v(x)=f(g(h(x)))$, 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 $f(x)={{x}^{3}}$, $g(x)=\ln x$ and $h(x)={{x}^{2}}+1$ in the above equation, we get $\dfrac{dv(x)}{dx}=\dfrac{d{{\ln }^{3}}({{x}^{2}}+1)}{d\ln ({{x}^{2}}+1)}\times \dfrac{d\ln ({{x}^{2}}+1)}{d({{x}^{2}}+1)}\times \dfrac{d}{dx}({{x}^{2}}+1).....\left( 6 \right)$.
We already know the value of $\dfrac{d\ln ({{x}^{2}}+1)}{d({{x}^{2}}+1)}$ and $\dfrac{d}{dx}({{x}^{2}}+1)$ using equation $(2)$ and $(3)$.
To find $\dfrac{d{{\ln }^{3}}({{x}^{2}}+1)}{d\ln ({{x}^{2}}+1)}$, let’s assume $z=\ln ({{x}^{2}}+1)$.
Thus, we have $\dfrac{d{{\ln }^{3}}({{x}^{2}}+1)}{d\ln ({{x}^{2}}+1)}=\dfrac{d({{z}^{3}})}{dz}$.
We know that differentiation of any function of the form $y=a{{x}^{n}}+b$ is $\dfrac{dy}{dx}=an{{x}^{n-1}}$.
Substituting $a=1,n=3,b=0$ in the above equation, we get $\dfrac{d{{\ln }^{3}}({{x}^{2}}+1)}{d\ln ({{x}^{2}}+1)}=\dfrac{d({{z}^{3}})}{dz}=3{{z}^{2}}=3{{\ln }^{2}}({{x}^{2}}+1).....\left( 7 \right)$.
Substituting equation $(2)$, $(3)$ and $(7)$ in equation $(6)$, we get

& \dfrac{dv(x)}{dx}=\dfrac{d{{\ln }^{3}}({{x}^{2}}+1)}{d\ln ({{x}^{2}}+1)}\times \dfrac{d\ln ({{x}^{2}}+1)}{d({{x}^{2}}+1)}\times \dfrac{d}{dx}({{x}^{2}}+1) \\\ & \dfrac{dv(x)}{dx}=\dfrac{3{{\ln }^{2}}({{x}^{2}}+1)}{{{x}^{2}}+1}\times 2x=\dfrac{6x{{\ln }^{2}}({{x}^{2}}+1)}{{{x}^{2}}+1}.....\left( 8 \right) \\\ \end{aligned}$$ Now, we have to find the derivative of the function $$u(x)=\dfrac{8x}{{{x}^{2}}+1}$$. We will use quotient rule of differentiation which states that if $$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)=8x$$ and $$\beta (x)={{x}^{2}}+1$$ in the above equation, we get $$\dfrac{du(x)}{dx}=\dfrac{({{x}^{2}}+1)\times \dfrac{d(8x)}{dx}-8x\times \dfrac{d({{x}^{2}}+1)}{dx}}{{{({{x}^{2}}+1)}^{2}}}.....\left( 9 \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=8,n=1,b=0$$ in the above equation, we get $$\dfrac{d(8x)}{dx}=8.....\left( 10 \right)$$. Substituting $$a=1,n=2,b=1$$ in the above equation, we get $$\dfrac{d({{x}^{2}}+1)}{dx}=2x.....\left( 11 \right)$$. Substituting equation $$(10)$$ and $$(11)$$ in equation $$(9)$$, we get $$\begin{aligned} & \dfrac{du(x)}{dx}=\dfrac{({{x}^{2}}+1)\times \dfrac{d(8x)}{dx}-8x\times \dfrac{d({{x}^{2}}+1)}{dx}}{{{({{x}^{2}}+1)}^{2}}} \\\ & \dfrac{du(x)}{dx}=\dfrac{({{x}^{2}}+1)\times 8-8x\times 2x}{{{({{x}^{2}}+1)}^{2}}}=\dfrac{8{{x}^{2}}+8-16{{x}^{2}}}{{{({{x}^{2}}+1)}^{2}}}=\dfrac{8(1-{{x}^{2}})}{{{({{x}^{2}}+1)}^{2}}}.....\left( 12 \right) \\\ \end{aligned}$$ Substituting equation $$(8)$$ and $$(12)$$ in equation $$(5)$$, we get $$\begin{aligned} & \dfrac{d}{dx}a(x)=v(x)\times \dfrac{d}{dx}u(x)+u(x)\dfrac{d}{dx}v(x) \\\ & \dfrac{d}{dx}a(x)={{\ln }^{3}}({{x}^{2}}+1)\times \dfrac{8(1-{{x}^{2}})}{{{({{x}^{2}}+1)}^{2}}}+\dfrac{8x}{{{x}^{2}}+1}\times \dfrac{6x{{\ln }^{2}}({{x}^{2}}+1)}{{{x}^{2}}+1} \\\ \end{aligned}$$ Further simplifying the equation by taking common from both terms, we get $$\dfrac{d}{dx}a(x)=\dfrac{8{{\ln }^{2}}({{x}^{2}}+1)}{{{({{x}^{2}}+1)}^{2}}}\left\\{ (1-{{x}^{2}})\ln ({{x}^{2}}+1)+6{{x}^{2}} \right\\}$$. Thus, we have $$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}a(x)=\dfrac{8{{\ln }^{2}}({{x}^{2}}+1)}{{{({{x}^{2}}+1)}^{2}}}\left\\{ (1-{{x}^{2}})\ln ({{x}^{2}}+1)+6{{x}^{2}} \right\\}$$. Hence, the second derivative of $$y={{\ln }^{4}}({{x}^{2}}+1)$$ is $$\dfrac{8{{\ln }^{2}}({{x}^{2}}+1)}{{{({{x}^{2}}+1)}^{2}}}\left\\{ (1-{{x}^{2}})\ln ({{x}^{2}}+1)+6{{x}^{2}} \right\\}$$. **Note:** It is necessary to use the chain rule of differentiation for composition of three 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.