Question

Consider the given expression,$f(x)={{\sin }^{-1}}\left( \sin x \right)$, then
a.$f'(\dfrac{3\pi }{4})=1$
b.$f'(\dfrac{5\pi }{4})=-1$
c.\[f'(\dfrac{\pi }{2})\] does not exist
d.\[f'(\pi )\] does not exist

Answer 1

Hint: Use the formula of Inverse Trigonometric Functions in its domain only as they have not included any particular domain is not included in the problem.
${{\sin }^{-1}}(\sin \theta )=\theta $ For $\dfrac{-\pi }{2}\le \theta \le \dfrac{\pi }{2}$ , to solve the given problem.

Complete step-by-step answer:

We will write the given equation first,
$f(x)={{\sin }^{-1}}\left( \sin x \right)$
As they have not included any domain so we can proceed with standard formula which is the key concept to solve the given problem,
Formula:
${{\sin }^{-1}}(\sin \theta )=\theta $
For $\dfrac{-\pi }{2}\le \theta \le \dfrac{\pi }{2}$
As particular domain is not included in the problem therefore the function will not change with domain and we can just us the formula with it domain only and there is no need to define the function at the points other than the domain therefore, we can use the above formula directly,
Therefore we can write given equation as,
$f(x)={{\sin }^{-1}}\left( \sin x \right)$
By using the formula written above we can write the above equation as,
$\therefore f(x)=x$
Now we will differentiate the above function with respect to ‘x’, to get the required answer.
$\therefore \dfrac{d}{dx}f(x)=\dfrac{d}{dx}\left( x \right)$
To solve further we should know the formula given below,
Formula:
$\dfrac{d}{dx}\left( x \right)=1$
By using above formula we can write,
$\therefore f'(x)=1$
As $f'(x)$ is a constant function, therefore at any value it is going to be constant.
Now, put the value of ‘x’ as $\dfrac{3\pi }{4}$to get the final answer,
$\therefore f'(\dfrac{3\pi }{4})=1$Is the only correct answer from all the four options.
Hence, the correct answer is Option (a) $f'(\dfrac{3\pi }{4})=1$

Note: Always check whether the domain is given or not, in this type of problems, as it can change your answer. As they have not included any particular domain here and hence we have defined the required function in its domain only as shown below,
${{\sin }^{-1}}(\sin \theta )=\theta $
For $\dfrac{-\pi }{2}\le \theta \le \dfrac{\pi }{2}$