Question

If\[f(x)=\ln (\ln x)\], then find the value of \[{{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}_{x=e}}\]

Answer 1

Hint: Assume a $f(x)=y$ to avoid confusion. Then differentiate using the rules of differentiation.

Complete step by step answer:
 The given expression is
\[f(x)=\ln (\ln x)\]
Let $f(x)=y$ , then the above expression can be written as,
$y=\ln (\ln x)$
Now we will find the first order derivative. So, differentiate the above expression with respect to $'x'$, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}(\ln (\ln x))$
Now we know, $\left( \ln u \right)'=\dfrac{1}{\ln u}(u)'$ , applying this formula, the above equation can be written as,
$\dfrac{dy}{dx}=\dfrac{1}{\ln x}\dfrac{d}{dx}\left( \ln x \right)$
Again applying the formula, $\left( \ln u \right)'=\dfrac{1}{\ln u}(u)'$, we get
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{1}{\ln x}\times \dfrac{1}{x}............(i) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{{{x}^{-1}}}{\ln x} \\
\end{align}$
Now we will find the second order derivative. So, differentiate the above expression with respect to $'x'$, we get
$\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{{{x}^{-1}}}{\ln x} \right)$
Now we know the quotient rule, i.e., \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{d}{dx}u-u\dfrac{d}{dx}v}{{{v}^{2}}}\], applying this formula in the above equation, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{\ln x\dfrac{d}{dx}\left( {{x}^{-1}} \right)-{{x}^{-1}}\dfrac{d}{dx}\left( \ln x \right)}{{{\left( \ln x \right)}^{2}}} \right)$
Now we know $\dfrac{d}{dx}({{u}^{n}})=n{{u}^{n-1}}\dfrac{d}{dx}(u)$ , applying this formula, the above equation becomes,
$\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{\ln x(-1){{x}^{-1-1}}-{{x}^{-1}}\dfrac{1}{x}}{{{\left( \ln x \right)}^{2}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{-\dfrac{\ln x}{{{x}^{2}}}-\dfrac{1}{{{x}^{2}}}}{{{\left( \ln x \right)}^{2}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{-\dfrac{\ln x+1}{{{x}^{2}}}}{{{\left( \ln x \right)}^{2}}} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( \dfrac{\ln x+1}{{{\left( x\ln x \right)}^{2}}} \right).........(ii) \\
\end{align}$
Substituting value from equation (i), we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( \ln x+1 \right)\dfrac{dy}{dx}$
Now we will find the third order derivative. So, differentiate the above expression with respect to $'x'$, we get
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=-\dfrac{d}{dx}\left( \left( \ln x+1 \right)\dfrac{dy}{dx} \right)$
We know the product rule as, \[\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u\], applying this formula in the above equation, we get
$\Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-\left[ \left( \ln x+1 \right)\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)+\left( \dfrac{dy}{dx} \right)\dfrac{d}{dx}\left( \ln x+1 \right) \right]$
Solving this using the above mentioned formulas, we get
$\Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-\left[ \left( \ln x+1 \right)\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)+\left( \dfrac{dy}{dx} \right)\left( \dfrac{1}{x} \right) \right]$
Now substituting back the values from equation (i) and (ii), we get
$\Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-\left[ \left( \ln x+1 \right)\left( -\left( \dfrac{\ln x+1}{{{\left( x\ln x \right)}^{2}}} \right) \right)+\left( \dfrac{1}{x\ln x} \right)\left( \dfrac{1}{x} \right) \right]$
Now opening the brackets, we get
$\begin{align}
  & \Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-\left[ -\left( \dfrac{{{\left( \ln x+1 \right)}^{2}}}{{{\left( x\ln x \right)}^{2}}} \right)+\left( \dfrac{1}{{{x}^{2}}\ln x} \right) \right] \\
 & \Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}={{\left( \dfrac{\ln x+1}{x\ln x} \right)}^{2}}-\dfrac{1}{{{x}^{2}}\ln x} \\
\end{align}$
Now we will substitute $'x=e'$, in the above expression, we get
$\Rightarrow {{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}_{x=e}}={{\left( \dfrac{\ln (e)+1}{e\ln (e)} \right)}^{2}}-\dfrac{1}{{{e}^{2}}\ln (e)}$
We know $\ln e=1$, substituting this value, we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}_{x=e}}={{\left( \dfrac{1+1}{e(1)} \right)}^{2}}-\dfrac{1}{{{e}^{2}}(1)} \\
 & \Rightarrow {{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}_{x=e}}={{\left( \dfrac{2}{e} \right)}^{2}}-\dfrac{1}{{{e}^{2}}} \\
 & \Rightarrow {{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}_{x=e}}=\dfrac{4-1}{{{e}^{2}}} \\
 & \Rightarrow {{\left( \dfrac{{{d}^{3}}y}{d{{x}^{3}}} \right)}_{x=e}}=\dfrac{3}{{{e}^{2}}} \\
\end{align}\]

Note: Another way of solving is to find the value of each degree of derivative. Then this can be substitute in the equation $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=-\left[ \left( \ln x+1 \right)\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)+\left( \dfrac{dy}{dx} \right)\left( \dfrac{1}{x} \right) \right]$.