Question

Find ‘$\dfrac{{dy}}{{dx}}$’ at $x = \dfrac{\pi }{4}$. If given that: $y = \left| {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right|$.
(a) Doesn’t exist
(b) $1$
(c) Cannot be determined
(d) None of these

Answer 1

Hint : The given problem revolves around the concepts of derivatives as well as algebraic theory respectively. As a result, derivating the given function with respect to ‘$x$’, substituting $x = \dfrac{\pi }{4}$ and then using the definition of modulus the desired solution is obtained.

Complete step-by-step answer :
Since, we have given that
$y = \left| {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right|$
Deriving the equation that is $y = \left| {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right|$ with respect to ‘$x$’, we get
$\dfrac{{dy}}{{dx}} = y' = \left| {\dfrac{d}{{dx}}\left[ {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right]} \right|$
As a result, solving the equation mathematically that is we know that derivation of ‘tan x’ is always ${\sec ^2}x$, we get
$\dfrac{{dy}}{{dx}} = y' = \left| { - {{\sec }^2}\left( {\dfrac{\pi }{4} - x} \right)} \right|$
Where,
Where, derivation of any term raised to other term is always $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ respectively.
But, since the given main function exists in the modulus sign
As a result, by its definition of modulus
Any function consisting in to the modulus sign has always two possible cases that is ‘positive’ and ‘negative’,
Hence, taking out the modulus sign, we get
$y = + \left[ { - {{\sec }^2}\left( {\dfrac{\pi }{4} - x} \right)} \right]$ At, $x \geqslant 0$
Or,
$y = - \left[ { - {{\sec }^2}\left( {\dfrac{\pi }{4} - x} \right)} \right]$ At, $x < 0$
Solving the equation algebraically, we get
$y = - {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)$ At, $x \geqslant 0$
Or,
$y = {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)$ At, $x < 0$
But,
Given that at, $x = \dfrac{\pi }{4}$, the equation becomes
$y = - {\sec ^2}\left( {\dfrac{\pi }{4} - \dfrac{\pi }{4}} \right)$ At, $x \geqslant 0$
Or,
$y = {\sec ^2}\left( {\dfrac{\pi }{4} - \dfrac{\pi }{4}} \right)$ At, $x < 0$
Hence, solving the equation mathematically, we get
$y = - {\sec ^2}0$ At, $x \geqslant 0$
Or,
$y = {\sec ^2}0$ At, $x < 0$
Now, since we know that trigonometric value at an angle ‘zero’ is $\sec {0^ \circ } = 1$, we get
$y = - 1$ At, $x \geqslant 0$
Or,
$y = 1$ At, $x < 0$
As a result, from the above equation it seems that the conditions of the modulus sign that is at $x \geqslant 0$ and $x < 0$ both contradicts with the solution calculated above i.e.
$\Rightarrow - 1 \ne > 0$ and $1 \ne < 0$
Hence, the solution does not exist for the given values respectively.
So, the correct answer is “Option a”.

Note : One must be able to know all the formulae/rules of derivatives such as $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, etc. to solve the respective solution particularly. Understand the definition of modulus which exists in two possible ways that is ‘positive’ and ‘negative’, so as to be sure of our final answer.