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Question: What is the derivative of, \(-\csc \left( x \right)\) ?...

What is the derivative of, −csc⁡(x)-\csc \left( x \right) ?

Explanation

Solution

We know that ‘csc’ is a term used for writing the “cosec” trigonometric function. Thus, our problem is basically to find the derivative of −cos⁡ec(x)-\cos ec\left( x \right) . We will use the chain rule to find the derivative of [−sin⁡(x)]−1{{\left[ -\sin \left( x \right) \right]}^{-1}} with respect to −sin⁡(x)-\sin \left( x \right) and then multiply it with the derivative of −sin⁡(x)-\sin \left( x \right) with respect to (x)(x). This will give us the required solution.

Complete step-by-step answer:
Let us first assign some terms that we are going to use later in our problem.
Let the given term on which we need to operate a differential be given by ‘y’ . Here, ‘y’ is given to us as:
⇒y=−csc⁡(x)\Rightarrow y=-\csc \left( x \right)
This can further be written as:
⇒y=[−sin⁡(x)]−1\Rightarrow y={{\left[ -\sin \left( x \right) \right]}^{-1}}
Then, we need to find the differential of ‘y’ with respect to ‘x’. This can be done as follows:
⇒dydx=d[−sin⁡(x)]−1dx\Rightarrow \dfrac{dy}{dx}=\dfrac{d{{\left[ -\sin \left( x \right) \right]}^{-1}}}{dx}

On applying chain rule, we can simplify the above equation as follows:
⇒dydx=d[−sin⁡(x)]−1d[−sin⁡(x)]×d[−sin⁡(x)]dx\Rightarrow \dfrac{dy}{dx}=\dfrac{d{{\left[ -\sin \left( x \right) \right]}^{-1}}}{d\left[ -\sin \left( x \right) \right]}\times \dfrac{d\left[ -\sin \left( x \right) \right]}{dx}
Let us name the above equation as (1)(1), so we have:
⇒dydx=d[−sin⁡(x)]−1d[−sin⁡(x)]×d[−sin⁡(x)]dx\Rightarrow \dfrac{dy}{dx}=\dfrac{d{{\left[ -\sin \left( x \right) \right]}^{-1}}}{d\left[ -\sin \left( x \right) \right]}\times \dfrac{d\left[ -\sin \left( x \right) \right]}{dx} ………(1)(1)

Using the formula for differential of a power, that is:
⇒d(θ)ndθ=n(θ)n−1\Rightarrow \dfrac{d{{\left( \theta \right)}^{n}}}{d\theta }=n{{\left( \theta \right)}^{n-1}}
And the formula for differential of a sine function, that is:
⇒d[sin⁡(θ)]dθ=cos⁡(θ)\Rightarrow \dfrac{d\left[ \sin \left( \theta \right) \right]}{d\theta }=\cos \left( \theta \right)

We get our equation as:
⇒dydx=(−1)[−sin⁡(x)]−1−1×d[−sin⁡(x)]dx ⇒dydx=−1sin⁡2x×d[−sin⁡(x)]dx ⇒dydx=−1sin⁡2x×(−cos⁡x) ⇒dydx=1sin⁡x×cos⁡xsin⁡x ∴dydx=csc⁡(x).cot⁡(x) \begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\left( -1 \right){{\left[ -\sin \left( x \right) \right]}^{-1-1}}\times \dfrac{d\left[ -\sin \left( x \right) \right]}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{{{\sin }^{2}}x}\times \dfrac{d\left[ -\sin \left( x \right) \right]}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{{{\sin }^{2}}x}\times \left( -\cos x \right) \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sin x}\times \dfrac{\cos x}{\sin x} \\\ & \therefore \dfrac{dy}{dx}=\csc \left( x \right).\cot \left( x \right) \\\ \end{aligned}
Thus, the final result comes out to be csc⁡(x).cot⁡(x)\csc \left( x \right).\cot \left( x \right)
Hence, the derivative of −csc⁡(x)-\csc \left( x \right) comes out to be csc⁡(x).cot⁡(x)\csc \left( x \right).\cot \left( x \right) .

Note: It is very important to know the meaning of terms like ‘csc’ as they are the short form of writing ‘cosec’, as these are some very common terms. Also, one should know the differential result of these trigonometric quantities as they are some very important results. Although they can be very easily derived, it is recommended to remember them thoroughly as it will save time spent in extra calculation.