Question

How do you find the derivative of \[\csc x\]?

Answer 1

Hint: In this question, we will use the concept of the quotient rule to find the derivative of the given trigonometric ratio. In this, first, we will write \[\csc x\] in $\sin x$ and then use the quotient rule for the derivative.

Complete step by step answer:
In this question, we have given the function \[\csc x\] and we need to determine its derivative.
First, we will write \[\csc x\] in terms of function as,
\[ \Rightarrow y = \csc x\]
As we know that $\csc x$ is the ratio of the hypotenuse and the perpendicular length of the right angle triangle while $\sin x$ is the ratio of the perpendicular and the hypotenuse, so $\csc x$ is the reciprocal of $\sin x$. From the above discussion we can write,
\[ \Rightarrow y = \dfrac{1}{{\sin x}}\]
For the quotient formula, let us assume,
\[ \Rightarrow y = \dfrac{u}{v}\]
Here, $u$ and $v$ are the function of $x$
Now, we will write the quotient formula for the derivative of the function as,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\]
Now, calculate the derivative of the given term by the quotient formula,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sin x\dfrac{{d\left( 1 \right)}}{{dx}} - \left( 1 \right)\dfrac{{d\left( {\sin x} \right)}}{{dx}}}}{{{{\left( {\sin x} \right)}^2}}}\]
Now, simplify the above equation as,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {\sin x} \right)\left( 0 \right) - \left( 1 \right)\left( {\cos x} \right)}}{{{{\sin }^2}x}}\]
After simplifying further, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\cos x}}{{{{\sin }^2}x}}.....\left( 1 \right)\]
As we know that, trigonometric ratio $\cot x$ is the ratio of the function $\cos x$ and $\sin x$ that is,
$ \Rightarrow \cot x = \dfrac{{\cos x}}{{\sin x}}$
So, by using the above term equation (1) will become,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \cot x\left( {\dfrac{1}{{\sin x}}} \right)\]
And as we know that $\csc x$ is the reciprocal of $\sin x$, so we can write
\[\therefore \dfrac{{dy}}{{dx}} = - \cot x\csc x\]

Therefore, the derivative of \[\csc x\] is \[ - \cot x\csc x\].

Note: As we know that the derivative of the constant term will be zero and the derivative of $\sin x$ is $\cos x$. We can use another method to calculate the derivative of $\csc x$ by using the identity to convert it into $\cot x$, but it will be complicated, so the quotient rule is best for this function.