Question

What is the derivative of ${\left( {\dfrac{1}{{\sin x}}} \right)^2}$?

Answer 1

We have to find the derivative of ${\left( {\dfrac{1}{{\sin x}}} \right)^2}$. We know that, inverse of sine is cosine. Hence, we can write $\left( {\dfrac{1}{{\sin x}}} \right)$ as $\cos ecx$. So, now we have to differentiate ${\left( {\cos ecx} \right)^2}$. To find the derivative of ${\left( {\cos ecx} \right)^2}$, we will be using the chain rule, which is $\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$

Complete step-by-step solution:
In this question, we have to find the derivative of ${\left( {\dfrac{1}{{\sin x}}} \right)^2}$.
Let the derivative of ${\left( {\dfrac{1}{{\sin x}}} \right)^2}$ be equal to x.
Therefore,
$x = \dfrac{d}{{dx}}{\left( {\dfrac{1}{{\sin x}}} \right)^2}$- - - - - - - - - (1)
Now, the above expression may look a little complicated, but it is quite simple.
Now, we know that 1 divided by $\sin x$ is equal to $\cos ecx$, as the inverse of $\sin x$ is $\cos ecx$.
$\Rightarrow \dfrac{1}{{\sin x}} = \cos ecx$.
Putting this value in equation (1), we get
$\Rightarrow x = \dfrac{d}{{dx}}{\left( {\cos ecx} \right)^2}$- - - - - - - - (2)
Now, to differentiate the above equation, we need to use the chain rule.
According to chain rule,
$\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$
Therefore, we have to differentiate $\cos ecx$ and power 2 as well.
Now, the derivative of $\cos ecx$ is $- \cos ecx \cdot \cot x$ and the derivative of $\cos e{c^2}x$ is $2\cos ecx$.
$\Rightarrow \dfrac{d}{{dx}}\left( {\cos ecx} \right) = - 2\cos ecx \cdot \cot x$
Therefore, equation (2) becomes,
$\Rightarrow x = - 2\cos ecx \cdot \cot x.\cos ecx \\\ \Rightarrow x = - 2\cos e{c^2}x \cdot \cot x \\\$
Hence, the derivative of ${\left( {\dfrac{1}{{\sin x}}} \right)^2}$ is $- 2\cos ec{x^2} \cdot \cot x$.

Note: We can also solve this question using a division method.
$\Rightarrow x = \dfrac{1}{{{{\sin }^2}x}}$
Now, the division rule is
$\Rightarrow \dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right) \cdot f'\left( x \right) - f\left( x \right) \cdot g'\left( x \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$
Here, $f\left( x \right) = 1$ and $g\left( x \right) = {\sin ^2}x$. Therefore putting these values in above equation, we get
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{{{{\sin }^2}x}}} \right) = \dfrac{{{{\sin }^2}x\dfrac{d}{{dx}}1 - 1\dfrac{d}{{dx}}{{\sin }^2}x}}{{{{\left( {{{\sin }^2}x} \right)}^2}}} \\\ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{{{{\sin }^2}x}}} \right) = \dfrac{{0 - 2\sin x\cos x}}{{{{\sin }^4}x}} = \dfrac{{ - 2\cos x}}{{{{\sin }^3}x}} \\\$
Now, we know that $\dfrac{{\cos x}}{{\sin x}} = \cot x$ and $\dfrac{1}{{\sin x}} = \cos ecx$.
Therefore, the above equation becomes,
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{{{{\sin }^2}x}}} \right) = \dfrac{{ - 2\cos x}}{{{{\sin }^2}x \cdot \sin x}} = - 2\cot x \cdot \cos e{c^2}x$