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.