Question

If $y=\log \left( \log x \right)$ , then $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ is equal to
(a) $\dfrac{-\left( 1+\log x \right)}{{{x}^{2}}\log x}$
(b) $\dfrac{\left( 1+\log x \right)}{{{x}^{2}}\log x}$
(c) $\dfrac{-\left( 1+\log x \right)}{{{\left( x\log x \right)}^{2}}}$
(d) $\dfrac{\left( 1+\log x \right)}{{{\left( {{x}^{2}}\log x \right)}^{2}}}$

Answer 1

Hint: Take exponential on both sides, and then simplify using the definition of logarithm that says ${{e}^{\log a}}=a$ . Differentiate both sides to obtain $\dfrac{dy}{dx}$ and then again differentiate to obtain $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ . Substitute and rearrange to obtain the desired result.

Complete step by step solution:
Here, we have a complicated function which needs to be differentiated. So, we can either use the chain rule of differentiation along with substitution, or we can convert this complicated function into a simple function using the concept of logarithms. We are going to use the second method here.
We have,
$y=\log \left( \log x \right)$
Let us take the exponentials on both sides. Now, we get
${{e}^{y}}={{e}^{\log \left( \log x \right)}}$
We know about the identity of logarithms, ${{e}^{\log a}}=a$ . Hence, we have
${{e}^{y}}=\log x...\left( i \right)$

Differentiating equation (i), we get
$\dfrac{d\left( {{e}^{y}} \right)}{dx}=\dfrac{d\left( \log x \right)}{dx}...(ii)$
We are well aware that $\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}\text{ and }\dfrac{d\left( \log x \right)}{dx}=\dfrac{1}{x}$ .
Using these two formulae in equation (ii), we get
${{e}^{y}}\dfrac{dy}{dx}=\dfrac{1}{x}...(iii)$
We can rewrite this equation as $\dfrac{dy}{dx}=\dfrac{1}{x{{e}^{y}}}...(iv)$
We need $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ , so we have to differentiate the equation again. Differentiating equation (iii), we get
$\dfrac{d}{dx}\left( {{e}^{y}}\dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{1}{x} \right)...\left( v \right)$
We know that $\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=-\dfrac{1}{{{x}^{2}}}$ .
We are also aware of the product rule of differentiation, which says $\dfrac{d}{dx}\left( u\cdot v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ .
Hence equation (v) becomes,
$\dfrac{d}{dx}\left( {{e}^{y}}\dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left( \dfrac{1}{x} \right)$
$\Rightarrow {{e}^{y}}\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)+\left( \dfrac{dy}{dx} \right)\left\{ \dfrac{d\left( {{e}^{y}} \right)}{dx} \right\}=-\dfrac{1}{{{x}^{2}}}$
We can simplify the above equation as
$\Rightarrow {{e}^{y}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\left( \dfrac{dy}{dx} \right)\left\{ {{e}^{y}}\dfrac{dy}{dx} \right\}=-\dfrac{1}{{{x}^{2}}}$
$\Rightarrow {{e}^{y}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+{{e}^{y}}{{\left( \dfrac{dy}{dx} \right)}^{2}}=-\dfrac{1}{{{x}^{2}}}...\left( vi \right)$
Putting the value $\dfrac{dy}{dx}=\dfrac{1}{x{{e}^{y}}}$ from equation (iv) into equation (vi), we get
${{e}^{y}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+{{e}^{y}}{{\left( \dfrac{1}{x{{e}^{y}}} \right)}^{2}}=-\dfrac{1}{{{x}^{2}}}$
On simplifying, we get
${{e}^{y}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{1}{{{x}^{2}}}-{{e}^{y}}{{\left( \dfrac{1}{x{{e}^{y}}} \right)}^{2}}$
Dividing both sides by ${{e}^{y}}$ , we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{1}{{{x}^{2}}{{e}^{y}}}-\dfrac{1}{{{x}^{2}}{{e}^{2y}}}$
Putting the value of ${{e}^{y}}$ from equation (i), we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{1}{{{x}^{2}}\log x}-\dfrac{1}{{{x}^{2}}{{\left( \log x \right)}^{2}}}$
We can simplify the RHS by taking LCM to get,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-\left( \log x+1 \right)}{{{x}^{2}}{{\left( \log x \right)}^{2}}}$
We can rewrite this as
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-\left( 1+\log x \right)}{{{\left( x\log x \right)}^{2}}}$

So, the correct answer is “Option c”.

Note: Alternatively, we can solve this problem using Chain Rule of Differentiation and some substitution, but that is a more complex approach for this problem. We must remember these formulae by heart and should take care in the calculation part, as any small silly mistake may lead to getting a different option, as the options are quite similar.