Question

If $y = {\tan ^{ - 1}}\left( {{{\log }_{(e{x^6})}}\left( {\dfrac{{{e^2}}}{{{x^3}}}} \right)} \right) + {\tan ^{ - 1}}\left( {{{\log }_{(e{x^{12}})}}\left( {\dfrac{{{e^4}}}{{{x^3}}}} \right)} \right)$, then $\dfrac{{dy}}{{dx}}$ is equal to
A. $3{\tan ^{ - 1}}\left( {\log x} \right)$
B. $0$
C. $\dfrac{1}{2}$
D. None of these

Answer 1

Hint: We know the base change properties of ${\log _a}b$ is given, we can write it as $\dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$
So here $\left( {{{\log }_{(e{x^6})}}\left( {\dfrac{{{e^2}}}{{{x^3}}}} \right)} \right)$ can be written as
$\dfrac{{{{\log }_e}\left( {\dfrac{{{e^2}}}{{{x^3}}}} \right)}}{{{{\log }_e}(e{x^6})}} = \dfrac{{{{\log }_e}{e^2} - {{\log }_e}{x^3}}}{{{{\log }_e}e + {{\log }_e}{x^6}}}$
And then solve the function.

Complete step-by-step answer:
Before solving this, we should know some log formulae:
$\log (ab) = \log a + \log b$
$\log (\dfrac{a}{b}) = \log a - \log b$
${\log _a}b$$ = $$\dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$ (base change properties)
Now we are given that
$y = {\tan ^{ - 1}}\left( {{{\log }_{(e{x^6})}}\left( {\dfrac{{{e^2}}}{{{x^3}}}} \right)} \right) + {\tan ^{ - 1}}\left( {{{\log }_{(e{x^{12}})}}\left( {\dfrac{{{e^4}}}{{{x^3}}}} \right)} \right)$
Now applying the base change formula, we get that
$y = {\tan ^{ - 1}}\dfrac{{{{\log }_e}\left( {\dfrac{{{e^2}}}{{{x^3}}}} \right)}}{{{{\log }_e}(e{x^6})}}$$ + {\tan ^{ - 1}}\dfrac{{{{\log }_e}\left( {\dfrac{{{e^4}}}{{{x^3}}}} \right)}}{{{{\log }_e}(e{x^{12}})}}$
As we know that
$\log (\dfrac{a}{b}) = \log a - \log b$
$\log (ab) = \log a + \log b$
${\log _a}{b^x} = x{\log _a}b$
We now know all the properties of log and by using them, we get that:
$y = {\tan ^{ - 1}}\left( {\dfrac{{{{\log }_e}{e^2} - {{\log }_e}{x^3}}}{{{{\log }_e}e + {{\log }_e}{x^6}}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{{{\log }_e}{e^4} - {{\log }_e}{x^3}}}{{{{\log }_e}e + {{\log }_e}{x^{12}}}}} \right)$
$y = {\tan ^{ - 1}}\left( {\dfrac{{2 - 3{{\log }_e}x}}{{1 + 6{{\log }_e}x}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{4 - 3{{\log }_e}x}}{{1 + 12{{\log }_e}x}}} \right)$
Also we know that
${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)$
So $y = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{2 - 3{{\log }_e}x}}{{1 + 6{{\log }_e}x}} + \dfrac{{4 - 3{{\log }_e}x}}{{1 + 12{{\log }_e}x}}}}{{1 - \dfrac{{2 - 3{{\log }_e}x}}{{1 + 6{{\log }_e}x}}.\dfrac{{4 - 3{{\log }_e}x}}{{1 + 12{{\log }_e}x}}}}} \right)$
$y = {\tan ^{ - 1}}\left( {\dfrac{{(2 - 3{{\log }_e}x)(1 + 12{{\log }_e}x) + (4 - 3{{\log }_e}x)(1 + 6{{\log }_e}x)}}{{(1 + 6{{\log }_e}x)((1 + 12{{\log }_e}x) - (3 - 2{{\log }_e}x)(4 - 3{{\log }_e}x)}}} \right)$
Upon simplification, we get that
$y = {\tan ^{ - 1}}\left( {\dfrac{{2 + 24{{\log }_e}x - 3{{\log }_e}x - 36{{({{\log }_e}x)}^2} + 4 + 24{{\log }_e}x - 3{{\log }_e}x - 18{{({{\log }_e}x)}^2}}}{{1 + 12{{\log }_e}x + 6{{\log }_e}x + 72{{({{\log }_e}x)}^2} - 8 + 6{{\log }_e}x + 12{{\log }_e}x - 9{{({{\log }_e}x)}^2}}}} \right)$
$y = {\tan ^{ - 1}}\left( {\dfrac{{6 + 42{{\log }_e}x - 54{{({{\log }_e}x)}^2}}}{{ - 7 + 36{{\log }_e}x + 63{{({{\log }_e}x)}^2}}}} \right)$
So $\dfrac{{dy}}{{dx}}$ of${\tan ^{ - 1}}(x)$ would be $\dfrac{1}{{1 + {x^2}}}$
$\dfrac{{dy}}{{dx}}$$ = \dfrac{1}{{1 + {{\left( {\dfrac{{6 + 42{{\log }_e}x - 54{{({{\log }_e}x)}^2}}}{{ - 7 + 36{{\log }_e}x + 63{{({{\log }_e}x)}^2}}}} \right)}^2}}}$
Now to solve this complete let us take ${\log _e}x = t$
So $\dfrac{{dy}}{{dx}}{|_t}$$ = \dfrac{1}{{1 + {{\left( {\dfrac{{6 + 42t - 54{t^2}}}{{63{t^2} + 36t - 7}}} \right)}^2}}}$
$ = \dfrac{{{{(63{t^2} + 36t - 7)}^2}}}{{{{(63{t^2} + 36t - 7)}^2} + {{(6 + 42t - 54{t^2})}^2}}}$
As we know that ${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca$
So we get that $\dfrac{{dy}}{{dx}}{|_t}$ as
$ = \dfrac{{3969{t^4} + 1269{t^2} + 49 + 4536{t^3} - 504t - 882{t^2}}}{{(3969{t^4} + 1269{t^2} + 49 + 4536{t^3} - 504t - 882{t^2}) + (2916{t^4} + 1764{t^2} + 36 - 4536{t^3} + 504t - 648{t^2})}}$$ = \dfrac{{3969{t^4} + 414{t^2} + 49 + 4536{t^3} - 504t}}{{6885{t^4} + 1530{t^2} + 85}}$
Where $t = {\log _e}x$
So no option is matching so none of these above options is the correct one
So it cannot be zero and it neither can be $3{\tan ^{ - 1}}(\log x)$ as its form is very complex.

Note: The derivative of the constant term is zero. Like if we are given $y = $$5$ then $\dfrac{{dy}}{{dx}}$ is always zero.
We must also know that $\dfrac{{dy}}{{dx}}$ of ${\tan ^{ - 1}}(x)$ would be $\dfrac{1}{{1 + {x^2}}}$