Question

Let a trigonometric function be given as \[y={{\tan }^{-1}}\left( \sec x+\tan x \right)\]. Then, \[\dfrac{dy}{dx}\] is equal to
(a) \[\dfrac{1}{4}\]
(b) \[\dfrac{1}{2}\]
(c) \[\dfrac{1}{\sec x+\tan x}\]
(d) \[\dfrac{1}{{{\sec }^{2}}x}\]
(e) \[\dfrac{1}{\tan x}\]

Answer 1

Hint: Consider the function of \[y\left( x \right)\] as a composite function and use the formula,
\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right)\times g'\left( x
\right)\], where \[y=f\left( g\left( x \right) \right)\] and \[f'\left( g\left( x \right) \right)\] means
differentiating \[f\left( g\left( x \right) \right)\] keeping \[g\left( x \right)\] constant and \[g'\left( x
\right)\] means differentiating \[g\left( x \right)\] with respect to x.

Complete step-by-step solution -
We are given with the function,
\[y={{\tan }^{-1}}\left( \sec x+\tan x \right)\]
Now we have to find the value of \[\dfrac{dy}{dx}\], which means that we have to differentiate y with
respect to x.
Now let us consider two functions \[f\left( x \right)\] and \[g\left( x \right)\], where \[f\left( x \right)\] be
\[{{\tan }^{-1}}x\] and \[g\left( x \right)\] be \[\left( \sec x+\tan x \right)\].
So, we can write,
\[y=f\left( g\left( x \right) \right)\]
Now, to differentiate y we have to use the identity.
\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right)\times g\left( x
\right)\]
Here, \[f'\left( g\left( x \right) \right)\] means that differentiating the function \[f\left( x \right)\] keeping
\[g\left( x \right)\] constant and \[g'\left( x \right)\] means differentiating only function \[g\left( x
\right)\].
So, to find \[\dfrac{d}{dx}\] of \[{{\tan }^{-1}}\left( \sec x+\tan x \right)\] we have to first know the
\[\dfrac{d}{dx}\] of some functions such as \[{{\tan }^{-1}}x\] and \[\left( \sec x+\tan x \right)\].
We know that,
\[\dfrac{d}{dx}\left( {{\tan }^{-1}}x \right)=\dfrac{1}{1+{{x}^{2}}}\]
And \[\dfrac{d}{dx}\left( \sec x+\tan x \right)\], which can be written as, \[\dfrac{d}{dx}\left( \sec x
\right)+\dfrac{d}{dx}\left( \tan x \right)\] or \[\sec x\tan x+{{\sec }^{2}}x\].
So, the \[\dfrac{d}{dx}\] of \[{{\tan }^{-1}}\left( \sec x+\tan x \right)\] is \[\dfrac{1\times \sec x\tan
x+{{\sec }^{2}}x}{1+{{\left( \sec x+\tan x \right)}^{2}}}\], which equals to \[\dfrac{\sec x\left( \sec x+\tan x
\right)}{1+{{\left( \sec x+\tan x \right)}^{2}}}\].
Now we will expand \[{{\left( \sec x+\tan x \right)}^{2}}\] and break into simpler terms. So, on using
identity, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\], we will break it into \[{{\sec }^{2}}x+{{\tan
}^{2}}x+2\sec x\tan x\], then we will use the identity, \[{{\tan }^{2}}x={{\sec }^{2}}x-1\]. So, we write
\[{{\left( \sec x+\tan x \right)}^{2}}\] as \[{{\sec }^{2}}x+\left( {{\sec }^{2}}x-1 \right)+2\sec x\tan x\] or
\[2{{\sec }^{2}}x-1+2\sec x\tan x\].
Now we can write the fraction as,
\[\dfrac{\sec x\left( \sec x+\tan x \right)}{1+2{{\sec }^{2}}x-1+2\sec x\tan x}\]
Hence on simplifying,
\[\dfrac{\sec x\left( \sec x+\tan x \right)}{2{{\sec }^{2}}x+2\sec x\tan x}\], which can be written as,
\[\dfrac{\sec x\left( \sec x+\tan x \right)}{2\sec x\left( \sec x+\tan x \right)}\]
So, on simplification we get the value of fraction as \[\dfrac{1}{2}\].
Hence, the value of \[\dfrac{dy}{dx}=\dfrac{1}{2}\].
So, the correct option is (b).

Note: Students can also change \[\sec x+\tan x\] in \[\sin x,\cos x\] terms and change into only \[\tan \] ratios but the solution is very big and tedious it’s preferred not to do it.