Question

If $y = {x^3}\log {\log _e}(1 + x)$ the $y''(0)$ equals:
A) $0$
B) $- 1$
C) $6{\log _e}2$
D) $6$

Answer 1

The question asks us to find the double differentiation of the function given in the question. Double differentiation also denoted by the ‘’ sign on the function is achieved by performing two differentiation successively on a function. We will first do the differentiation two times and then put the value of the variable equal to zero to get the value of $y''(0)$.

Formula used:
$\dfrac{d}{{dx}}(\log x) = \dfrac{1}{x}$
$\dfrac{d}{{dx}}(1 + x) = 1$
$\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$

Complete step by step answer:
We are given the function:
$y = {x^3}\log {\log _e}(1 + x)$
we will first differentiate two times then put the variable equal to zero.
We will differentiate it by using the product formula:
The product formula is given by:
$(xy)' = xy' + yx'$
We will take ${x^3}\& \log {\log _3}(1 + x)$ as the two functions.
$\Rightarrow y' = 3{x^2}\log {\log _e}(1 + x) + \dfrac{{{x^3}}}{{1 + x}} \times \dfrac{1}{{{{\log }_e}(1 + x)}}$
The above differentiation is done using the product rule. First term contains the differentiation of one function while the other is kept intact, and vice versa is done in the second term. See clearly that the part $\log {\log _e}(1 + x)$ is a composite function and is differentiated by the chain rule in above differentiation.
The function will now be differentiated again to find the value of $y''$
This time there will be many functions that need to be differentiated:
$\Rightarrow y'' = (3{x^2}\log {\log _e}(1 + x))' + \left( {\dfrac{{{x^3}}}{{1 + x}} \times \dfrac{1}{{{{\log }_e}(1 + x)}}} \right)'$
The two terms and their differentiation will be done individually.
First and the second terms below are the differentiation of the first term given above using the product rule, the other three terms are the differentiation of the second term given above.
$\Rightarrow y'' = 6x\log {\log _e}(1 + x) + \dfrac{{3{x^2}}}{{{{\log }_e}(1 + x)}} \times \dfrac{1}{{1 + x}} + \dfrac{{{x^3}}}{{(1 + {x^2}){{\log }_e}(1 + x)}} + \dfrac{{{x^3}}}{{{{(1 + x)}^2}}} \times \dfrac{1}{{{{[1 + {{\log }_e}(1 + x)]}^2}}} + \dfrac{{3{x^2}}}{{(1 + x){{\log }_e}(1 + x)}}$
Now we will put the variable in the function as $0$
We get:
$\Rightarrow y''(0) = 6 \times 0\log {\log _e}(1 + 0) + \dfrac{{3 \times {0^2}}}{{{{\log }_e}(1 + 0)}} \times \dfrac{1}{{1 + 0}} + \dfrac{{{0^3}}}{{(1 + {0^2}){{\log }_e}(1 + 0)}} + \dfrac{{{0^3}}}{{{{(1 + 0)}^2}}} \times \dfrac{1}{{{{[1 + {{\log }_e}(1 + 0)]}^2}}} + \dfrac{{3 \times {0^2}}}{{(1 + 0){{\log }_e}(1 + 0)}} \\\$
$\Rightarrow y''(0) = 0$
Hence, option (A) is correct.

Note:
We have used the method of chain rule to differentiate the function. The chain rule is quite a handy method when it comes to differentiating nested (composite) functions. Remember that we keep on differentiating the function until the last function is reached.
$f(g(x)) = f'(g(x)) \times g'(x)$
This in simpler terms means that first differentiate the outer function keeping the inner one intact. Then multiply the inner function’s differentiation to the previous result.