Question

Find the value of \[\dfrac{d}{{dx}}\sqrt {\dfrac{{1 - \sin 2x}}{{1 + \sin 2x}}} \] .
A. \[{\sec ^2}x\]
B. \[ - {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)\]
C. \[{\sec ^2}\left( {\dfrac{\pi }{4} + x} \right)\]
D. \[{\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)\]

Answer 1

Hint:
While watching the trigonometric equation inside the square root it looks very complex hence our first step will be to simplify the terms using the basic trigonometric formula and then we will be differentiating the term using chain rule.



Formula Used:
\[\sin 2x = 2\sin x\cos x\],
\[{\cos ^2}x + {\sin ^2}x = 1\] ,
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\],
\[\dfrac{d}{{dx}}[f(g(x))] = f\prime (g(x))g\prime (x)\]



Complete step-by-step answer:
By using the Double Angle Property we will solve the trigonometric equation inside square root.
Then \[\sin 2x\] can be written as \[2\sin x\cos x\] .
\[\dfrac{d}{{dx}}\sqrt {\dfrac{{1 - \sin 2x}}{{1 + \sin 2x}}} \]
\[ \Rightarrow \sqrt {\dfrac{{1 - 2\sin x\cos x}}{{1 + 2\sin x\cos x}}} \]
By using the formula \[{\cos ^2}x + {\sin ^2}x = 1\] we will substitute the value of 1 inside the square root.
\[ \Rightarrow \sqrt {\dfrac{{{{\cos }^2}x + {{\sin }^2}x - 2\sin x\cos x}}{{{{\cos }^2}x + {{\sin }^2}x + 2\sin x\cos x}}} \]
By applying the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] we can simplify the terms.
\[ \Rightarrow \sqrt {\dfrac{{{{\left( {\cos x - \sin x} \right)}^2}}}{{{{\left( {\cos x + \sin x} \right)}^2}}}} \]

\[ \Rightarrow \sqrt {{{\left( {\dfrac{{\left( {\cos x - \sin x} \right)}}{{\left( {\cos x + \sin x} \right)}}} \right)}^2}} \]
\[ \Rightarrow \left( {\dfrac{{\left( {\cos x - \sin x} \right)}}{{\left( {\cos x + \sin x} \right)}}} \right)\]
Dividing Numerator and Denominator both by \[\cos x\] to make the terms more simpler,
\[ \Rightarrow \left( {\dfrac{{\left( {\dfrac{{\cos x}}{{\cos x}}} \right) - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{{\cos x}}{{\cos x}}} \right) + \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}} \right)\]
 \[ \Rightarrow \left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)\]
According to the property we can write \[1\] as \[\tan \dfrac{\pi }{4}\]
\[ \Rightarrow \left( {\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan x}}{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan x}}} \right)\]
According to the formula \[ \Rightarrow \left( {\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan x}}{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan x}}} \right)\] can be written as \[\tan \left( {\dfrac{\pi }{4} - x} \right)\]
\[ \Rightarrow \tan \left( {\dfrac{\pi }{4} - x} \right)\]
We have got the inner trigonometric values in the simplest form and now we will be differentiating it,
 \[\dfrac{d}{{dx}}\left( {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right)\]
By Applying chain rule we will be differentiating the term.
\[\dfrac{d}{{dx}}[f(g(x))] = f\prime (g(x))g\prime (x)\]
Where \[f\left( x \right) = \tan x\] and \[g\left( x \right) = \left( {\dfrac{\pi }{4} - x} \right)\]
Applying the formula and differentiating the term we will get ,
\[\dfrac{d}{{dx}}\left( {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right)\]
Differentiation of \[\tan x = {\sec ^2}x\]
\[ \Rightarrow {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right).\dfrac{d}{{dx}}\left( {\dfrac{\pi }{4} - x} \right)\]
Differentiation of \[\dfrac{d}{{dx}}\left( {\dfrac{\pi }{4} - x} \right) = \left( { - 1} \right)\]
\[ \Rightarrow - {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)\]
Hence the answer is (B) which is\[ - {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)\].



Note:
Students generally make mistakes while using the trigonometric formula as well while using the differentiation via chain rule hence students must be very careful while doing the differentiation .