Solve the given expression, \[\dfrac{d(cosecx)}{dx}\] .

Hint: We know that, \[\text{cosecx=}\dfrac{\text{1}}{\text{sinx}}={{(sinx)}^{-1}}\] . We have the term \[\operatorname{sinx}\] in the numerator of the expression \[\dfrac{d{{(sinx)}^{-1}}}{dx}\] . Now using chain rule we can write \[\dfrac{d{{(sinx)}^{-1}}}{dx}\] as \[\dfrac{d{{(sinx)}^{-1}}}{d\sin x}\times \dfrac{d\sin x}{dx}\] . We know that, \[\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}\] and \[\dfrac{d\sin x}{dx}=\cos x\] . Now, using these formulas we can solve further.

Complete step-by-step answer:
According to the question, we have to solve \[\dfrac{d(cosecx)}{dx}\] ……………………(1)
Here, our function to differentiate is \[\text{cosecx}\] .
We know the relation between sine function and cosec function. That is, cosec is reciprocal of sine function.
\[\sin x=\dfrac{1}{\cos ecx}={{(sinx)}^{-1}}\] …………………(2)
Using equation (1) and equation (2), we get
\[\dfrac{d(cosecx)}{dx}=\dfrac{d{{(sinx)}^{-1}}}{dx}\] …………………(3)
In equation (3), we can see that direct differentiation is complex. We need to simplify it in simpler form.
Using chain rule, we can simplify it.
\[\dfrac{d{{(sinx)}^{-1}}}{d\sin x}\times \dfrac{d\sin x}{dx}\] ……………………….(4)
We know the formula,
\[\dfrac{d\sin x}{dx}=\cos x\] …………………….(5)
From equation (4), and equation (5), we get
\[\dfrac{d{{(sinx)}^{-1}}}{d\sin x}\times \dfrac{d\sin x}{dx}\]
\[\text{=}\dfrac{\text{d(sinx}{{\text{)}}^{\text{-1}}}}{\text{dsinx}}\text{ }\!\!\times\!\!\text{ cosx}\] …………………………(6)
Replacing \[\text{sinx}\] by y in the above equation(6), we get
\[\dfrac{d{{y}^{-1}}}{dy}\times \cos x\] …………………..(7)
We know the formula that,
\[\dfrac{d{{y}^{n}}}{dy}=n{{y}^{n-1}}\]
Putting n=-1 in the above equation, we get
\[\dfrac{d{{y}^{-1}}}{dy}=-{{y}^{-1-1}}=-{{y}^{-2}}\] …………………….(8)
From equation (7) and equation (8), we get
\[\dfrac{d{{y}^{-1}}}{dy}\times \cos x\]
\[=-{{y}^{-2}}.\cos x\] …………………(9)
We had replaced \[\text{sinx}\] by y. So here, we are replacing y by \[\text{sinx}\] .
Now, replacing y by \[\text{sinx}\] in equation (9), we get
\[=-{{y}^{-2}}.\cos x\]
\[=-{{(sinx)}^{-2}}\cos x\]
\[=-\dfrac{\cos x}{{{(sinx)}^{2}}}\] …………………..(10)
We know that, \[\text{cosecx=}\dfrac{\text{1}}{\text{sinx}}\] and \[\dfrac{\cos x}{\sin x}=\cot x\] .
Transforming equation (10), we get
\[=-\dfrac{\cos x}{{{(sinx)}^{2}}}\]
\[=-\dfrac{\cos x}{(sinx)(sinx)}\]
\[=-cotx.cosecx\]
Hence, \[\dfrac{d(cosecx)}{dx}=-cotx.cosecx\] .

Note: In chain rule one may think as \[\dfrac{d{{(sinx)}^{-1}}}{dx}=\dfrac{d{{(sinx)}^{-1}}}{d\cos x}\times \dfrac{d\cos x}{dx}\] . If we do so then our differentiation will become complex to solve. So, we can’t solve this question by this approach. We can see that the numerator of the expression \[\dfrac{d{{(sinx)}^{-1}}}{dx}\] has the term \[\operatorname{sinx}\] . So, in chain rule we have to use \[\operatorname{sinx}\] in order to make the differentiation easy to solve.