Question

Find the value of \[{10^{ - x\tan x}}\left[ {\dfrac{d}{{dx}}{{10}^{x\tan x}}} \right]\] .

Answer 1

Hint: This question is based on some basic concepts and formulas of trigonometry. Our first concept is that we have to solve the bracket then only we can deal with the outer expression. Inside the bracket we have given a derivative term so first of all keep in mind and try the derivative formulas that is \[{a^x}\] in this case but the power is also mixed up with the trigonometric function so multiplication rules for this. After the completion of the bracket and derivative part we are left with the outer expression. Now the term inside the bracket and outside the bracket are opposite in sign that means they can cancel each other. So by doing this we can easily get our final answer.

Complete step-by-step answer:
The given expression is \[{10^{ - x\tan x}}\left[ {\dfrac{d}{{dx}}{{10}^{x\tan x}}} \right]\] . ------- (i)
 So, to find the value of this expression first we will have to solve the term inside the square bracket i.e. , \[\dfrac{d}{{dx}}{10^{x\tan x}}\] .
Let \[y = {10^{x\tan x}}\] ------------- (ii)
By taking log on both sides we get
\[\log y = \log \left( {{{10}^{x\tan x}}} \right)\]
As we know that \[\log {m^n} = n\log m\] . Therefore by using this property the above equation becomes
\[\log y = x\tan x\log 10\]
Now differentiate with respect to \[x\] . And to differentiate \[x\tan x\] we need to apply the multiplication rule of differentiation by taking \[\log 10\] out because \[\log 10\] is a constant term. So,
\[\dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \log 10\left[ {x\dfrac{d}{{dx}}\left( {\tan x} \right) + \tan x\dfrac{d}{{dx}}\left( 1 \right)} \right]\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = y\log 10\left[ {x\dfrac{d}{{dx}}\left( {\tan x} \right) + \tan x} \right]\]
From equation (ii) we have
\[ \Rightarrow \dfrac{{dy}}{{dx}} = {10^{x\tan x}}\log 10\left[ {x{{\sec }^2}x + \tan x} \right]\] -------- (iii)
Put the value of equation (iii) in equation (i)
\[ \Rightarrow {10^{ - x\tan x}}\left[ {{{10}^{x\tan x}}\log 10\left( {x{{\sec }^2}x + \tan x} \right)} \right]\]
Shifting the negative power to the denominator
\[ \Rightarrow \dfrac{1}{{{{10}^{x\tan x}}}}{10^{x\tan x}}\left[ {\log 10\left( {x{{\sec }^2}x + \tan x} \right)} \right]\]
\[ \Rightarrow \log 10\left( {x{{\sec }^2}x + \tan x} \right)\]
Hence, \[{10^{ - x\tan x}}\left[ {\dfrac{d}{{dx}}{{10}^{x\tan x}}} \right]\] is equal to \[\log 10\left( {x{{\sec }^2}x + \tan x} \right)\] .
So, the correct answer is “\[\log 10\left( {x{{\sec }^2}x + \tan x} \right)\]”.

Note: Differentiation of \[\tan x\] is \[{\sec ^2}x\] . Always try to solve the expression given inside the bracket. Multiplication rule of differentiation is \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}\left( v \right) + v\dfrac{d}{{dx}}\left( u \right)\] . Keep in mind that \[\log {m^n} = n\log m\] . In \[\log y = x\tan x\log 10\] you have to differentiate \[x\tan x\] by using \[uv\] rule only. \[\log 10\] is a constant so make sure you don’t make such a mistake by differentiating the expression by taking \[x\tan x\] as \[u\] and \[\log 10\] as \[v\] .