Question

If $y = {\log _a}\left( x \right) + {\log _x}\left( a \right) + {\log _x}\left( x \right) + {\log _a}\left( a \right)$, then $\dfrac{{dy}}{{dx}}$is equal to
(A) $\dfrac{1}{x} + x\log \left( a \right)$
(B) $\dfrac{{\log \left( a \right)}}{x} + \dfrac{x}{{\log \left( a \right)}}$
(C) $x\log \left( a \right)$
(D) $\dfrac{1}{{x\log \left( a \right)}} - \dfrac{{\log \left( a \right)}}{{x{{\left( {\log \left( x \right)} \right)}^2}}}$

Answer 1

Hint : We solve this question by first applying the log property that ${\log _m}\left( n \right) = \dfrac{{\log (n)}}{{\log (m)}}$ to solve the y as y is complex. To make it simple we first simplify y by applying log property and then taking differentiating both sides with respect to x so as to find $\dfrac{{dy}}{{dx}}$. We should remember that $\dfrac{d}{{dx}}\left( {\log (x)} \right) = \dfrac{1}{x}$ and derivative of constant term is zero and derivative of $\dfrac{d}{{dx}}\left( {c{x^n}} \right) = cn{\left( x \right)^{\left( {n - 1} \right)}}$ and the basic chain rule of derivative which states that derivative of a function of function is given by first differentiating the dependent function multiply derivative of independent function.$\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = \dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right)\left[ {\dfrac{d}{{dx}}\left( {g\left( x \right)} \right)} \right]$

Complete step-by-step answer :
We are given, $y = {\log _a}\left( x \right) + {\log _x}\left( a \right) + {\log _x}\left( x \right) + {\log _a}\left( a \right)$ and we have to find $\dfrac{{dy}}{{dx}}$.
First we will apply the log property that ${\log _m}\left( n \right) = \dfrac{{\log (n)}}{{\log (m)}}$ to simplify y
So we get,
$y = \dfrac{{\log \left( x \right)}}{{\log \left( a \right)}} + \dfrac{{\log \left( a \right)}}{{\log \left( x \right)}} + \dfrac{{\log \left( x \right)}}{{\log \left( x \right)}} + \dfrac{{\log \left( a \right)}}{{\log \left( a \right)}}$
Now, on further simplifications, we get,
$y = \dfrac{{\log \left( x \right)}}{{\log \left( a \right)}} + \dfrac{{\log \left( a \right)}}{{\log \left( x \right)}} + 1 + 1$
$y = \dfrac{{\log \left( x \right)}}{{\log \left( a \right)}} + \dfrac{{\log \left( a \right)}}{{\log \left( x \right)}} + 2$
As we know that, derivative of constant function is zero, derivative of $\log \left( x \right)$ is $\dfrac{1}{x}$ and derivative of $\dfrac{1}{x}$is $\dfrac{{ - 1}}{{{x^2}}}$
Now, differentiating both sides with respect to x using the chain rule of differentiation $\dfrac{{d\left( {f\left( {g\left( x \right)} \right)} \right)}}{{dx}} = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$, we get,
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{x\log \left( a \right)}} + \dfrac{{\left( { - 1} \right)\log \left( a \right)}}{{x{{\left( {\log \left( x \right)} \right)}^2}}}$
We get, $\dfrac{d}{{dx}}\left( {\dfrac{1}{{\log (x)}}} \right) = \dfrac{{ - 1}}{{x{{\left( {\log \left( x \right)} \right)}^2}}}$, and we know that $\dfrac{d}{{dx}}\left( {c{x^n}} \right) = cn{\left( x \right)^{\left( {n - 1} \right)}}$ which implies that,
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{x\log \left( a \right)}} - \dfrac{{\log \left( a \right)}}{{x{{\left( {\log \left( x \right)} \right)}^2}}}$
Hence, option (D) is the correct answer.
So, the correct answer is “Option D”.

Note : The main thing while doing this question is to first simplify y so as to calculate its derivative easily. Keep in mind the log property that ${\log _m}\left( n \right) = \dfrac{{\log (n)}}{{\log (m)}}$. Keep in mind the basic differentiation formulas like, derivative of constant function is zero, derivative of $\log \left( x \right)$ is $\dfrac{1}{x}$ and derivative of $\dfrac{1}{x}$ is $\dfrac{{ - 1}}{{{x^2}}}$ and the basic chain rule which states that $\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = \dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right)\left[ {\dfrac{d}{{dx}}\left( {g\left( x \right)} \right)} \right]$. Take care while doing the calculations.