Question

How do you find the derivative of $\csc x$?

Answer 1

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.