Question

If $y=\ln \left( \operatorname{cosec}x-\cot x \right)$ then, $\dfrac{dy}{dx}$ equals to,
A. $\operatorname{cosec}x+\cot x$
B. $\cot x$
C. $\sec x+\tan x$
D. $\operatorname{cosec}x$

Answer 1

Hint: For solving this question, we should know about the differentiation formulas of trigonometric function. In this type of question, we always calculate the first order derivative of the function and the first order derivative is calculated by the chain rule in this type of question.

Complete step by step answer:
The function given to us in the question is,
$f\left( x \right)=y=\ln \left( \operatorname{cosec}x-\cot x \right)$
The first order derivative of the given function is,
$f'\left( x \right)=\dfrac{dy}{dx}$
We will differentiate $f\left( x \right)$ with respect to $x$ one time. By the chain rule, we get,
$f'\left( x \right)=\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$
In the chain rule, we consider any term of the function as ‘u’ and the set to the differentiation in the given form,
$\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$
Here, in our question, $u=\operatorname{cosec}x-\cot x$, so we will substitute that value. So, we will get,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{1}{\operatorname{cosec}x-\cot x}.\dfrac{d\left( \operatorname{cosec}x-\cot x \right)}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{-\operatorname{cosec}x.\cot x+{{\operatorname{cosec}}^{2}}x}{\operatorname{cosec}x-\cot x}\ldots \ldots \ldots \left( A \right) \\
\end{align}$
By further solving our equation (A), we find that,
$\dfrac{dy}{dx}=\dfrac{-\operatorname{cosec}x\left( \cot x-\operatorname{cosec}x \right)}{\left( \operatorname{cosec}x-\cot x \right)}$
We know that we can also write this as,
$\dfrac{dy}{dx}=\dfrac{-\operatorname{cosec}x\left( \cot x-\operatorname{cosec}x \right)}{-\left( \cot x-\operatorname{cosec}x \right)}$
We can see that there are common terms in the numerator and the denominator that will divide each other. So, they cancel out and the remaining will become the solution of our question. So, we have,
$\dfrac{dy}{dx}=\operatorname{cosec}x$
So, the first derivative is $\dfrac{dy}{dx}=\operatorname{cosec}x$ for the given function.

So, the correct answer is “Option D”.

Note: In these types of questions the chain rule is very useful and we should use the chain rule for solving this. While considering the terms of the function for the chain rule, we must notice that term with denoting variable and while taking values again in the equation, we must be careful and should take the same value.