Question

How do you find the second derivative of $\ln \left( \dfrac{x+1}{x-1} \right)$ ?

Answer 1

Hint: We recall chain rule and quotient rule of differentiation. We use the logarithmic identity of quotient $\log \left( \dfrac{a}{b} \right)=\log a-\log b$ and find the first derivative using chain rule that is $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}$and then second derivative using quotient rule $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$.

Complete step-by-step answer:
We know that if the real valued functions $f\left( x \right),g\left( x \right)$ are defined within sets $f:A\to B$ and $g:B\to C$ then the composite function from A to C is defend as $g\left( f\left( x \right) \right)$ within sets $gof:A\to C$. If we denote $g\left( f\left( x \right) \right)=y$ and $f\left( x \right)=u$ then we can differentiate the composite function using chain rule as
\[\dfrac{d}{dx}g\left( f\left( x \right) \right)=\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}\]
 We also know from quotient rule of differentiation that if $u\left( x \right),v\left( x \right)$ are two real valued function with the condition $v\left( x \right)\ne 0$ then we can differentiate their quotient as
\[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]
We are given the function $\ln \left( \dfrac{x+1}{x-1} \right)$ to differentiate. Let use the logarithmic identity of quotient $\log \left( \dfrac{a}{b} \right)=\log a-\log b$ for $a=x+1,b=x-1$ to have
\[\ln \left( \dfrac{x+1}{x-1} \right)=\ln \left( x+1 \right)-\ln \left( x-1 \right)\]
Let us denote $y=,\ln \left( x+1 \right)-\ln \left( x-1 \right)w=\ln \left( x+1 \right),z=\ln \left( x-1 \right)$. So we have
\[y=w-z\]
We differentiate both sides of above equation to have
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{dw}{dx}-\dfrac{dz}{dx}\]
We use chain rule taking $u=x+1,v=x-1$ to have
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{dw}{du}\cdot \dfrac{du}{dx}-\dfrac{dw}{dv}\cdot \dfrac{dv}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{d\left( x+1 \right)}\ln \left( x+1 \right)\cdot \dfrac{d}{dx}\left( x+1 \right)-\dfrac{d}{d\left( x-1 \right)}\ln \left( x-1 \right)\cdot \dfrac{d}{dx}\left( x-1 \right) \\
\end{align}\]
We use the standard differentiation of logarithmic function $\dfrac{d}{dt}\log \left( t \right)=\dfrac{1}{t}$ for $t=x+1,t=x-1$ in the above step to have
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x+1}\cdot 1-\dfrac{1}{x-1}\cdot 1 \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{x+1}-\dfrac{1}{x-1} \\
\end{align}\]
We find the second derivative differentiating the first derivative again with respect to $x$. We have;
\[\Rightarrow \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{1}{x+1}-\dfrac{1}{x-1} \right)\]
We use sum rule of differentiation in the above step to have
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{1}{x+1} \right)-\dfrac{d}{dx}\left( \dfrac{1}{x-1} \right)\]
We use the quotient rule of differentiation for $u=1,v=x+1$ in the first term and $u=1,v=x-1$ in the second term of the right hand side. We have
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{1}{x+1} \right)-\dfrac{d}{dx}\left( \dfrac{1}{x-1} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( x+1 \right)\dfrac{d}{dx}1-1\cdot \dfrac{d}{dx}\left( x+1 \right)}{{{\left( x+1 \right)}^{2}}}-\dfrac{\left( x-1 \right)\dfrac{d}{dx}1-1\cdot \dfrac{d}{dx}\left( x-1 \right)}{{{\left( x-1 \right)}^{2}}} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( x+1 \right)0-1\cdot 1}{{{\left( x+1 \right)}^{2}}}-\dfrac{\left( x-1 \right)0-1\cdot 1}{{{\left( x-1 \right)}^{2}}} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{1}{{{\left( x+1 \right)}^{2}}}+\dfrac{1}{{{\left( x-1 \right)}^{2}}} \\
\end{align}\]
We add the fractional expression in the right hand side to have
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{\left( x+1 \right)}^{2}}-{{\left( x-1 \right)}^{2}}}{{{\left( x+1 \right)}^{2}}{{\left( x-1 \right)}^{2}}} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{\left( x+1 \right)}^{2}}-{{\left( x-1 \right)}^{2}}}{{{\left( \left( x+1 \right)\left( x-1 \right) \right)}^{2}}} \\
\end{align}\]
We use the algebraic identity ${{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}=4ab$ in the numerator and the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ for $a=x,b=1$ in the denominator of above step to have the second derivative ;
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{4\cdot x\cdot 1}{{{\left( {{x}^{2}}-1 \right)}^{2}}}=\dfrac{4x}{{{\left( {{x}^{2}}-1 \right)}^{2}}}\]

Note: We know that logarithmic function is only defined for positive real numbers. So we have$\dfrac{x+1}{x-1}\ne 0$. So we have$x\ne -1,1$. So our second derivative is well defined because${{x}^{2}}-1\ne 0$. We can differentiate without using the quotient rule with $\dfrac{d}{dt}\left( \dfrac{1}{t} \right)=-\dfrac{1}{{{t}^{2}}}$ for $t=x-1,x+1$. If $f\left( x \right),g\left( x \right)$ are real valued functions then the sum rule is given by $\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)$.