Question

If $f\left( x \right) = \left[ {\tan \dfrac{\pi }{4} + \tan x} \right]\left[ {\tan \dfrac{\pi }{4} + \tan \left( {\dfrac{\pi }{4} - x} \right)} \right]$ and $g\left( x \right) = x\left( {x + 1} \right)$ the value of $g'\left[ {f\left( x \right)} \right]$ is equal to
A. $\dfrac{1}{{1 + {x^2}}}$
B. $4$
C. $5$
D. $\dfrac{{2x}}{{1 + {x^2}}}$

Answer 1

Here we are two equations containing trigonometric values and we are asked to find the value of. Before getting into the question, we should solve and simplify the given equations as much as possible we can. Then, we need to find the first derivative of $g\left( x \right) = {x^2} + x$. And then, we shall calculate the value $g'\left[ {f\left( x \right)} \right]$. Formula to be used:
a) $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
b) $\tan \dfrac{\pi }{4} = 1$
c) The power rule for differentiation is as follows.
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

Complete step-by-step answer:
It is given that $f\left( x \right) = \left[ {\tan \dfrac{\pi }{4} + \tan x} \right]\left[ {\tan \dfrac{\pi }{4} + \tan \left( {\dfrac{\pi }{4} - x} \right)} \right]$
First, we shall solve the above-given equation.
$f\left( x \right) = \left[ {\tan \dfrac{\pi }{4} + \tan x} \right]\left[ {\tan \dfrac{\pi }{4} + \tan \left( {\dfrac{\pi }{4} - x} \right)} \right]$
$\Rightarrow \left[ {1 + \tan x} \right]\left[ {1 + \tan \left( {\dfrac{\pi }{4} - x} \right)} \right]$ (Here we applied $\tan \dfrac{\pi }{4} = 1$)
$\Rightarrow \left[ {1 + \tan x} \right]\left[ {1 + \dfrac{{\tan \dfrac{\pi }{4} - \tan x}}{{1 + \tan \dfrac{\pi }{4}\tan x}}} \right]$
(Since $\tan \left( {\dfrac{\pi }{4} - x} \right)$ is in the format $\tan \left( {A - B} \right)$, we have used $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$where $A = \dfrac{\pi }{4}$ and $B = 1$)
$\Rightarrow \left[ {1 + \tan x} \right]\left[ {1 + \dfrac{{1 - \tan x}}{{1 + 1 \times \tan x}}} \right]$ (Here we applied $\tan \dfrac{\pi }{4} = 1$)
$\Rightarrow \left[ {1 + \tan x} \right]\left[ {1 + \dfrac{{1 - \tan x}}{{1 + \tan x}}} \right]$
$\Rightarrow \left[ {1 + \tan x} \right]\left[ {\dfrac{{1 + \tan x + 1 - \tan x}}{{1 + \tan x}}} \right]$
$\Rightarrow \left[ {1 + \tan x} \right]\left[ {\dfrac{2}{{1 + \tan x}}} \right]$
$\Rightarrow 2$
So, we have simplified the given equation and we get $f\left( x \right) = 2$ .
We are also given another equation $g\left( x \right) = x\left( {x + 1} \right)$.
Now, we also need to simplify the above equation.
$g\left( x \right) = x\left( {x + 1} \right)$
$\Rightarrow {x^2} + x$
So, we have simplified the given equation and we get $g\left( x \right) = {x^2} + x$ .
We are asked to calculate $g'\left[ {f\left( x \right)} \right]$.
Before that, we need to find the derivative of $g\left( x \right)$ .
$\dfrac{d}{{dx}}\left( {g\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {{x^2} + x} \right)$
$\Rightarrow g'\left( x \right) = 2{x^{2 - 1}} + 1 \times {x^{1 - 1}}$ (Now we have applied the power rule for differentiation, $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$)
$\Rightarrow g'\left( x \right) = 2x + {x^0}$ (We know that the value of anything power zero always equals to one)
$\Rightarrow g'\left( x \right) = 2x + 1$
We are asked to calculate $g'\left[ {f\left( x \right)} \right]$.
We have $f\left( x \right) = 2$and $g'\left( x \right) = 2x + 1$ .
Thus, $g'\left[ {f\left( x \right)} \right] = 2 \times 2 + 1$
$g'\left[ {f\left( x \right)} \right] = 4 + 1$
$g'\left[ {f\left( x \right)} \right] = 5$
Hence, we get $g'\left[ {f\left( x \right)} \right] = 5$ and the option $C$ is correct.

So, the correct answer is “Option C”.

Note: Whenever our question involves trigonometry, we should be able to obtain the required formula and trigonometric identities for the given problem. Since we are asked to calculate the first derivation, we need to use the required formula for differentiation. And here, we have applied the power rule of differentiation.