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,

\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.