Question

If $y = {\log _5}\left( {{{\log }_7}x} \right)$, find $\dfrac{{dy}}{{dx}}$

Answer 1

To solve this question, we will use the chain rule and also use some basic logarithmic properties. Chain rule states that, suppose f is a real valued function which is a composite of three functions u, v and w, i.e. f = (w o u) o v. If $t = v\left( x \right)$ and $s = u\left( t \right)$, then $\dfrac{{df}}{{dt}} = \dfrac{{d\left( {wou} \right)}}{{dt}}.\dfrac{{dt}}{{dx}} = \dfrac{{dw}}{{ds}}.\dfrac{{ds}}{{dt}}.\dfrac{{dt}}{{dx}}$

Complete step-by-step answer :
Given that
$y = {\log _5}\left( {{{\log }_7}x} \right)$
Let $u = {\log _7}x$, then,
$y = {\log _5}u$ ……. (i)
According to the chain rule,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$ ……… (ii)
Differentiating equation (i) both sides with respect to u, we get
$\Rightarrow \dfrac{{dy}}{{du}} = \dfrac{d}{{du}}\left( {{{\log }_5}u} \right)$,
Now, using the change of base rule,
$\Rightarrow \dfrac{{dy}}{{du}} = \dfrac{d}{{du}}\left( {\dfrac{{{{\log }_e}u}}{{{{\log }_e}5}}} \right)$
$\Rightarrow \dfrac{{dy}}{{du}} = \dfrac{1}{{u{{\log }_e}5}}$
Putting the value of u,
$\Rightarrow \dfrac{{dy}}{{du}} = \dfrac{1}{{{{\log }_7}x{{\log }_e}5}}$
Now, we have
$u = {\log _7}x$
Using the change of base rule, we can write this as,
$u = \dfrac{{{{\log }_e}x}}{{{{\log }_e}7}}$
Differentiating both sides with respect to x, we will get
$\dfrac{{du}}{{dx}} = \dfrac{1}{{x{{\log }_e}7}}$
Now putting the values of $\dfrac{{dy}}{{du}}$ and $\dfrac{{du}}{{dx}}$ in equation (ii), we will get
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\log }_7}x{{\log }_e}5}} \times \dfrac{1}{{x{{\log }_e}7}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\log }_7}x{{\log }_e}5x{{\log }_e}7}}$
Hence, we can say that if $y = {\log _5}\left( {{{\log }_7}x} \right)$, then $\dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\log }_7}x{{\log }_e}5x{{\log }_e}7}}$

Note : Whenever we ask this type of questions, we have to remember the basic rule of differentiation and logarithm. According to the change of base rule in logarithm, ${\log _b}a$ can be written as $\dfrac{{{{\log }_x}a}}{{{{\log }_x}b}}$, where you can choose to change the logarithm to any base x. through this, we will get the answer.