Question

How do you find the derivative of $y=\dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)}$ ?

Answer 1

Hint: To get the derivative of $y=\dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)}$ with respect to x, we will use the formula of $\dfrac{dy}{dx}=\dfrac{d\left( \dfrac{u}{v} \right)}{dx}$ , where $u=\cos \left( x \right)$ , $v=\text{cosec}\left( x \right)$ and $\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{v.\dfrac{du}{dx}-u.\dfrac{dv}{dx}}{{{v}^{2}}}$ . After getting the derivative of $\cos \left( x \right)$ , $\text{cosec}\left( x \right)$ and using it in the formula then simplifying it, we will get the derivative of $y=\dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)}$ with respect to $x$ .

Complete step by step solution:
The given equation is $y=\dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)}$ that is in the form of $\dfrac{u}{v}$ , where we will get the values after comparison as $u=\cos \left( x \right)$ and $v=\text{cosec}\left( x \right)$.
We will use the formula of derivative of division for getting the derivative of the given equation which is:
$\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\dfrac{v.\dfrac{du}{dx}-u.\dfrac{dv}{dx}}{{{v}^{2}}}$
Since, we got the values of $u$ and $v$ , we will put these values in the above formula that will be as:
$\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\dfrac{\text{cosec}\left( x \right).\dfrac{d\left[ \cos \left( x \right) \right]}{dx}-\cos \left( x \right).\dfrac{d\left[ \text{cosec}\left( x \right) \right]}{dx}}{{{\left[ \text{cosec}\left( x \right) \right]}^{2}}}$
Since, the derivative of $\cos \left( x \right)$ is $\left[ -\sin \left( x \right) \right]$ and the derivative of $\text{cosec}\left( x \right)$ is $\left[ -\text{cosec}\left( x \right).\cot \left( x \right) \right]$ , we will use it in the above equation. So, the equation will be now as:
$\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\dfrac{\text{cosec}\left( x \right).\left[ -\sin \left( x \right) \right]-\cos \left( x \right).\left[ -\text{cosec}\left( x \right).\cot \left( x \right) \right]}{{{\left[ \text{cosec}\left( x \right) \right]}^{2}}}$
Now, we will simplify the equation step by step,
$\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\dfrac{\text{cosec}\left( x \right).\left[ -\sin \left( x \right) \right]-\cos \left( x \right).\left[ -\text{cosec}\left( x \right).\cot \left( x \right) \right]}{{{\left[ \text{cosec}\left( x \right) \right]}^{2}}}$
$\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\dfrac{-\sin \left( x \right).\text{cosec}\left( x \right)+\cos \left( x \right)\text{.cosec}\left( x \right).\cot \left( x \right)}{{{\left[ \text{cosec}\left( x \right) \right]}^{2}}}$
Since, we know that $\text{cosec}\left( x \right)$ is equal to $\dfrac{1}{\sin \left( x \right)}$ and $\cot \left( x \right)$ is equal to $\dfrac{\cos \left( x \right)}{\sin \left( x \right)}$ . So, here we will use them in the above equation so that above equation will be as:
\[\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\dfrac{-\sin \left( x \right).\dfrac{1}{\sin \left( x \right)}+\cos \left( x \right)\text{.}\dfrac{1}{\sin \left( x \right)}.\dfrac{\cos \left( x \right)}{\sin \left( x \right)}}{{{\left[ \dfrac{1}{\sin \left( x \right)} \right]}^{2}}}\]
Again we will simplify the above equation, so the process would be as:
\[\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\left[ -\sin \left( x \right).\dfrac{1}{\sin \left( x \right)}+\cos \left( x \right)\text{.}\dfrac{1}{\sin \left( x \right)}.\dfrac{\cos \left( x \right)}{\sin \left( x \right)} \right]\times {{\left[ \sin \left( x \right) \right]}^{2}}\]
\[\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\left\{ -1+{{\left[ \dfrac{\cos \left( x \right)}{\sin \left( x \right)} \right]}^{2}} \right\}\times {{\left[ \sin \left( x \right) \right]}^{2}}\]
\[\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\left\{ -1\times {{\left[ \sin \left( x \right) \right]}^{2}}+{{\left[ \dfrac{\cos \left( x \right)}{\sin \left( x \right)} \right]}^{2}}\times {{\left[ \sin \left( x \right) \right]}^{2}} \right\}\]
\[\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\left\{ -{{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right) \right\}\]
The above equation can be written as:
\[\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\left\{ {{\cos }^{2}}\left( x \right)-{{\sin }^{2}}\left( x \right) \right\}\]
In the above equation we got the derivative of the given equation as \[\left\{ {{\cos }^{2}}\left( x \right)-{{\sin }^{2}}\left( x \right) \right\}\]but \[\left\{ {{\cos }^{2}}\left( x \right)-{{\sin }^{2}}\left( x \right) \right\}\] is equal to \[\cos \left( 2x \right)\] . So, the above equation can be written as:
\[\Rightarrow \dfrac{d\left( \dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)} \right)}{dx}=\cos \left( 2x \right)\]
Hence, the derivative of $y=\dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)}$ with respect to $x$ is \[\dfrac{dy}{dx}=\cos \left( 2x \right)\] .

Note:
For finding the derivative of $y=\dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)}$ with respect to $x$, we can use another method that is in terms of multiplication. Since, we know that $\sin \left( x \right)$ is equal to $\dfrac{1}{\text{cosec}\left( x \right)}$ , we can make the changes in the given question with help of it. So, we can write the given equation as $y=\sin \left( x \right).\cos \left( x \right)$ . Now, we can use the derivative formula of multiplication that is:
 $\dfrac{d\left( u.v \right)}{dx}=u.\dfrac{dv}{dx}+\dfrac{du}{dx}.v$ , where we will use$u=\sin \left( x \right)$ and $v=\cos \left( x \right)$ in the equation. Since, The derivative of $\sin \left( x \right)$ is $\cos \left( x \right)$ and the derivative of $\cos \left( x \right)$ is $\left[ -\sin \left( x \right) \right]$. After applying these derivatives in the in the process solution and then simplifying it, we will get the derivative of the equation $y=\dfrac{\cos \left( x \right)}{\text{cosec}\left( x \right)}$ with respect to $x$:
\[\dfrac{dy}{dx}=\cos \left( 2x \right)\]