Question

How do you find the derivative of $\dfrac{1}{{\sin x}}$ ?

Answer 1

Hint: We have to find the differentiation of the function $\dfrac{1}{{\sin x}}$. Take 1 as the first function and $\sin x$ as the second function and differentiate by using the quotient rule. The rule which is to be used to solve this question is –
$\dfrac{{dy}}{{dx}}\left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}$
where, $f\left( x \right)$ is the first function and $g\left( x \right)$ is the second function.

Complete Step by Step Solution:
From the question, we are given with the function $\dfrac{1}{{\sin x}}$. In this function we can clearly see that this function is of the two differentiable functions. So, we can use the quotient rule of differentiation to solve this question.
In calculus, the quotient rule can be defined as the method for finding the differentiation of that function which is the ratio of two differentiable functions. Let $y = \dfrac{{f\left( x \right)}}{{g\left( x \right)}}$ , where both the functions $f$ and $g$ are differentiable and $g\left( x \right) \ne 0$. Then, by using the quotient rule the derivative of $y$ function can be done as –
$\dfrac{{dy}}{{dx}} = \dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}} \cdots \left( 1 \right)$
So, we have the function $\dfrac{1}{{\sin x}}$ .
Therefore, let $y = \dfrac{1}{{\sin x}}$ , $f\left( x \right) = 1$ and $g\left( x \right) = \sin x$
Putting the values of $y,f\left( x \right)$ and $g\left( x \right)$ from the above in the equation (1), we get –
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{{\sin x}}} \right) = \dfrac{{\sin x\dfrac{d}{{dx}}\left( 1 \right) - 1\dfrac{d}{{dx}}\left( {\sin x} \right)}}{{{{\left( {\sin x} \right)}^2}}}$
We know that the differentiation of any constant value is always 0, so, doing differentiation of 1, which is constant in the above equation, we get –
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{{\sin x}}} \right) = \dfrac{{\sin x \times 0 - \dfrac{d}{{dx}}\left( {\sin x} \right)}}{{{{\sin }^2}x}}$
We also know that the differentiation of $\sin x$ with respect to $x$ is $\cos x$ . So, doing the differentiation of $\sin x$ with respect to $x$ in the above equation, we get –
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{{\sin x}}} \right) = - \dfrac{{\cos x}}{{{{\sin }^2}x}}$
The above differentiation of the function $\dfrac{1}{{\sin x}}$ can also be simplified, so that it becomes easy. Therefore, it can also be written as –
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{{\sin x}}} \right) = - \dfrac{{\cos x}}{{\sin x}} \times \dfrac{1}{{\sin x}}$
Now, we know that the ratio of $\cos x$ and $\sin x$ is known as $\cot x$ which is opposite of $\tan x$ and the inverse of $\sin x$ is known as $\cos ecx$. So, using these to simplify the differentiation, we get –
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{{\sin x}}} \right) = - \cot \left( x \right)\cos ec\left( x \right)$

Hence, $ - \cot \left( x \right)\cos ec\left( x \right)$ is the required derivative of the function $\dfrac{1}{{\sin x}}$.

Note:
The function given in the question, $\dfrac{1}{{\sin x}}$ ,can also be written as $\cos ecx$ as it is the inverse of $\sin x$ and then, we can directly get our answer without using the quotient rule which is $ - \cot \left( x \right)\cos ec\left( x \right)$ as this is the differentiation of $\cos ecx$.
Always try to simplify your answer as much as you can because it gives the good impression on your answer.