Question

If $y = \sqrt {\dfrac{{1 + \tan x}}{{1 - \tan x}}}$ then $\dfrac{{dy}}{{dx}}$ is equal to –
A.$\dfrac{1}{2}\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} {\sec ^2}\left( {\dfrac{\pi }{4} + x} \right)$

B.$\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} {\sec ^2}\left( {\dfrac{\pi }{4} + x} \right)$

C.$\dfrac{1}{2}\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \sec \left( {\dfrac{\pi }{4} + x} \right)$

D.None of these

Answer 1

Hint : Here, we find the first order of the derivative and by using the chain rule of the composite function. Also, apply the laws of power and exponent and different trigonometric relations while simplification of the equation.

Complete step-by-step answer :
Take the given expression –
$y = \sqrt {\dfrac{{1 + \tan x}}{{1 - \tan x}}}$
Take derivative with respect to “x” on both the sides of the equation-
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\sqrt {\dfrac{{1 + \tan x}}{{1 - \tan x}}} } \right)$
Here first of all apply ${x^n}$ formula in the above equation and then apply $\left( {\dfrac{u}{v}} \right)$ rule.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}{\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)^{\dfrac{1}{2} - 1}}\dfrac{d}{{dx}}\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)$
Simplify the above equation –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}{\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)^{ - \dfrac{1}{2}}}\dfrac{d}{{dx}}\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)$
Apply law of the inverse of the power and exponent rule-
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\dfrac{d}{{dx}}\left( {\dfrac{{1 + \tan x}}{{1 - \tan x}}} \right)$
Apply division of two terms on the other part of the right hand side of the equation –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{{(1 - \tan x)\dfrac{{d(1 + \tan x)}}{{dx}} - (1 + \tan x)\dfrac{{d(1 - \tan x)}}{{dx}}}}{{{{(1 - \tan x)}^2}}}} \right)$
Place the derivative in the above equation –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{{(1 - \tan x){{\sec }^2}x + (1 + \tan x){{\sec }^2}x}}{{{{(1 - \tan x)}^2}}}} \right)$
Open the brackets and simplify the numerator –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{{{{\sec }^2}x - \tan x{{\sec }^2}x + {{\sec }^2}x + \tan x{{\sec }^2}x}}{{{{(1 - \tan x)}^2}}}} \right)$
Make the pair of like terms –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{{\underline {{{\sec }^2}x + {{\sec }^2}x} \underline { - \tan x{{\sec }^2}x + \tan x{{\sec }^2}x} }}{{{{(1 - \tan x)}^2}}}} \right)$
Like terms with same values and opposite sign, cancel each other –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{{2{{\sec }^2}x}}{{{{(1 - \tan x)}^2}}}} \right)$
Open the square of the function in the denominator –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{{2{{\sec }^2}x}}{{1 - 2\tan x + \tan {x^2}}}} \right)$
We know that $1 + {\tan ^2}x = {\sec ^2}x$ in the above equation –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{{2{{\sec }^2}x}}{{{{\sec }^2}x - 2\tan x}}} \right)$
Take ${\sec ^2}x$ common from both the numerator and the denominator and remove them since the common multiple from the numerator and the denominator cancel each other.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{1}{{\dfrac{1}{2} - \dfrac{{2\tan x}}{{2{{\sec }^2}x}}}}} \right)$
We know that secant is the inverse function of cosine substitute in the above equation –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{1}{{\dfrac{1}{2} - \sin x\cos x}}} \right)$
Substitute the value for $1 = {\sin ^2}x + {\cos ^2}x$, also multiply and divide by $2$ to make a perfect square.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{1}{{\dfrac{1}{2}({{\sin }^2}x + {{\cos }^2}x) - 2\dfrac{{\sin x}}{{\sqrt 2 }}\dfrac{{\cos x}}{{\sqrt 2 }}}}} \right)$
We can re-write as the square of two terms –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{1}{{{{\left( { - \dfrac{{\sin x}}{{\sqrt 2 }} + \dfrac{{\cos x}}{{\sqrt 2 }}} \right)}^2}}}} \right)$
It can be re-written as –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {\dfrac{1}{{\left( {{{\cos }^2}(\dfrac{\pi }{4} + x)} \right)}}} \right)$
Cosine is the inverse of the secant function. Therefore, it can be re-written as –
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left( {\sqrt {\dfrac{{1 - \tan x}}{{1 + \tan x}}} } \right)\left( {{{\sec }^2}(\dfrac{\pi }{4} + x)} \right)$ is the required solution.
So, the correct answer is “Option A”.

Note : Always remember the basic trigonometric and the derivative formulas for the accurate and efficient solution. Remember the relation between the trigonometric functions to substitute the values in and during the solution. Also, go through the chain rule of the derivatives and the first, second and third order of the derivative.