Question

The value of \[\dfrac{d}{{dx}}{\cosh ^{ - 1}}(\sec x)\] is equal to:
A. \[\sec x\]
B. \[\tan x\]
C. \[\sin x\]
D. \[{\mathop{\rm cosec}\nolimits} x\]

Answer 1

Hint:
The given question will be solved by a chain rule method. First we will write the differentiation of \[\cosh ^{ - 1}x \] where \[x =\sec x\]. Then multiply the differentiation of \[\sec x\].



Formula Used:
\[\dfrac{d}{{dx}}{\cosh ^{ - 1}}x = \dfrac{1}{{\sqrt {{x^2} - 1} }}\]
\[\dfrac{d}{{dx}}{\sec x = {\sec x\tan x}}\]
\[{\tan ^2}{\rm{ x = }}{\sec ^2}x - 1\]




Complete step-by-step answer:
We have been given the function \[\dfrac{d}{{dx}}{\cosh ^{ - 1}}(\sec x)\].

Let \[y = {\cosh ^{ - 1}}(\sec x)\]
Differentiating y with respect to x, we get,
\[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {\left( {{{\sec }^2}x - 1} \right)} }}\sec x\tan x\]
Now we will apply the property of \[{\tan ^2}{\rm{ x = }}{\sec ^2}x - 1\]
\[ = \dfrac{{\sec x\tan x}}{{\sqrt {{{\tan }^2}x} }}\]
Now take the square root of \[{\tan ^2}x\]
\[ = \dfrac{{\sec x\tan x}}{{\tan x}}\]
\[ = \sec x\]
Hence, option A is correct.

Additional information
Chain rule is used to differentiate composite functions. The differentiate of a composite
 \[f\left( {g\left( x \right)} \right)\] is \[\frac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)\]
To apply chain rule, we need to identify the outer function and inner function. Then we will write the derivative of outer function and multiply the differentiation of inner function.



Note:
Students often make common mistakes to find the differentiation of \[\frac{d}{{dx}}{\cosh ^{ - 1}}(\sec x)\]. They do not use chain rule to solve the question and treat \[\sec x\] as \[x\]. The correct way is first to differentiate \[{\cosh ^{ - 1}}(\sec x)\] and multiply the differentiation of \[\sec x\] with it.