Question

If ${\text{y = sec}}\left( {{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}} \right)$, then $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$at x = 1 is equal to?
$
  {\text{A}}{\text{. }}\dfrac{1}{2} \\
  {\text{B}}{\text{. 1}} \\
  {\text{C}}{\text{. }}\sqrt 2 \\
  {\text{D}}{\text{. }}\dfrac{1}{{\sqrt 2 }} \\
$

Answer 1

Hint: In order to find the value of $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ at x = 1, we differentiate the given function ‘y’ with respect to x and then substitute the value x = 1 in the obtained to find the answer. The first order differentials of trigonometric and inverse trigonometric functions are used while simplification.

Complete step-by-step answer:
Given Data,
${\text{y = sec}}\left( {{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}} \right)$
Let us differentiate the given function ‘y’ with respect to ‘x’,
$ \Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{sec}}\left( {{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}} \right)} \right)$
We know the derivative of sec x is tan x and the derivative of ${\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}$ is $\dfrac{1}{{1 + {{\text{x}}^2}}}$
$ \Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = sec}}\left( {{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}} \right){\text{tan}}\left( {{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}} \right) \times \dfrac{1}{{1 + {{\text{x}}^2}}}$
We know that the inverse of a function itself gives us the value of the variable in the function, i.e. $f\left( {{{\text{f}}^{ - 1}}\left( {\text{x}} \right)} \right) = {\text{x}}$
$ \Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{\text{ = sec}}\left( {{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}} \right) \times \dfrac{{\text{x}}}{{1 + {{\text{x}}^2}}}$
Now let us compute $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ at x = 1,
$ \Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{{\text{|}}_{{\text{x = 1}}}}{\text{ = sec}}\left( {{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}} \right) \times \dfrac{{\text{x}}}{{1 + {{\text{x}}^2}}}$
We know ${\text{ta}}{{\text{n}}^{ - 1}}\left( 1 \right) = 45^\circ {\text{ and sec 45}}^\circ {\text{ = }}\sqrt 2 $
$ \Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{{\text{|}}_{{\text{x = 1}}}}{\text{ = sec}}\left( {45^\circ } \right) \times \dfrac{{\text{x}}}{{1 + {{\text{x}}^2}}}$
$ \Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{{\text{|}}_{{\text{x = 1}}}}{\text{ = }}\sqrt 2 \times \dfrac{1}{2}$
$ \Rightarrow \dfrac{{{\text{dy}}}}{{{\text{dx}}}}{{\text{|}}_{{\text{x = 1}}}}{\text{ = }}\dfrac{1}{{\sqrt 2 }}$

Therefore if ${\text{y = sec}}\left( {{\text{ta}}{{\text{n}}^{ - 1}}{\text{x}}} \right)$, then $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ at x = 1 is equal to $\dfrac{1}{{\sqrt 2 }}$. So, Option D is the correct answer.

Note: In order to solve this type of questions the key is to know the concept of differentiation and the general formulae of few first order differentials of trigonometric and inverse trigonometric functions. The term $\dfrac{{{\text{dy}}}}{{{\text{dx}}}}$ means we differentiate the function represented by ‘y’ with respect to the variable ‘x’. The value of a function at a variable value is the value of the function when the variable in it takes the given value. The formula of differentials of trigonometric functions have to be memorized to easily solve problems of this type.