Question

Find the derivative of $\dfrac{x+\cos x}{\text{tan}x}$.

Answer 1

Hint:To differentiate the function given is a question; we will have to use quotient rule of differentiation and sum rule of differentiation.

Complete step-by-step answer:
The function given in question for which we have to find derivative is in the form of $\dfrac{\text{f}\left( x \right)}{g\left( x \right)}$ where $f\left( x \right)=x+\cos x$ and $g\left( x \right)=\tan x$ . To differentiate a function of the form $\dfrac{\text{f}\left( x \right)}{g\left( x \right)}$ , we will use the quotient rule of differentiation. The quotient rule of differentiate says that the derivative of the function $\dfrac{\text{f}\left( x \right)}{g\left( x \right)}$ is calculated as shown below:
$\dfrac{d}{dx}\left[ \dfrac{f\left( x \right)}{g\left( x \right)} \right]=\dfrac{g\left( x \right)\left( \dfrac{d}{dx}\left[ f\left( x \right) \right] \right)-f\left( x \right)\left( \dfrac{d}{\text{dx}}\left[ g\left( x \right) \right] \right)}{{{\left[ g\left( x \right) \right]}^{2}}}$ .
In our question, we have to find the difference of $\dfrac{x+\cos x}{\tan x}$ . In our case, $f\left( x \right)=x+\cos x$ and $g\left( x \right)=\tan x$ .
Therefore, after applying quotient rule of differentiation, we get:
$\dfrac{\text{d}}{dx}\left( \dfrac{x+\cos x}{\tan x} \right)=\dfrac{\tan x\left[ \dfrac{d}{dx}\left( x+\cos x \right) \right]-\left( x+\cos x \right)\left[ \dfrac{d}{dx}\left( \tan x \right) \right]}{{{\left( \tan x \right)}^{2}}}$ …….(i)
In the equation (1), we have to find the derivative of terms $\left( x+\cos x \right)$ and $\left( \tan x \right)$ , the derivative of $\left( x+\cos x \right)$can be calculated by sum rule of differentiate. The sum rule of differentiate says that the differentiation of the function $f\left( x \right)=q\left( x \right)+r\left( x \right)$ is given as:
$\dfrac{d}{dx}\left[ P\left( x \right) \right]=\dfrac{d}{dx}\left[ q\left( x \right)+r\left( x \right) \right]=\dfrac{d}{dx}\left[ q\left( x \right) \right]+\dfrac{d}{dx}\left[ r\left( x \right) \right]$ .
Using the above rule, the differentiation of $\left( x+\cos x \right)$ will be:
$\dfrac{\text{d}}{dx}\left[ x+\cos x \right]=\dfrac{d}{dx}\left[ x \right]+\dfrac{d}{dx}\left[ \cos x \right]$ .
Now, we know that $\dfrac{\text{d}}{dx}\left[ x \right]=1$ and $\dfrac{d}{dx}\left[ \cos x \right]=-\sin x.$ we will put these values in above equation. After doing this we will get:
$\dfrac{d}{dx}\left[ x+\cos x \right]=1-\sin x.......(ii)$
We also know that $\dfrac{\text{d}}{dx}\left[ \tan x \right]={{\sec }^{2}}x...........(iii)$
Now, we will put the value of required derivatives from equation (ii) and (iii) into equation (i). After doing this we will get:
$\Rightarrow \dfrac{\text{d}}{dx}\left[ \dfrac{x+\cos x}{\tan x} \right]=\dfrac{\tan x\left[ 1-\sin x \right]-\left( x+\cos x \right)\left( se{{c}^{2}}x \right)}{{{\tan }^{2}}x}$ .
We can also write the above equation as:
$\Rightarrow \dfrac{d}{dx}\left[ \dfrac{x+\cos x}{\tan x} \right]=\dfrac{\left( 1-\sin x \right)}{\tan x}-\dfrac{\left( x+\cos x \right)\left( {{\sec }^{2}}x \right)}{{{\tan }^{2}}x}$ .
$\Rightarrow \dfrac{d}{dx}\left[ \dfrac{x+\cos x}{\tan x} \right]=\cot x-\cos x-\dfrac{x{{\sec }^{2}}x}{{{\tan }^{2}}x}-\dfrac{\sec x}{{{\tan }^{2}}x}$ .
$\Rightarrow \dfrac{d}{dx}\left[ \dfrac{x+\cos x}{\tan x} \right]=\cot x-\cos x-x\cos e{{c}^{2}}x-\cot x\cos ecx$ .
Hence, this is our required solution.

Note: We have calculated a general function of the derivative of $\left( \dfrac{x+\cos x}{\tan x} \right)$ . If we will put different values of x as the above calculated derivative, we will get the values of the derivative at particular points. The important thing to note here is that the function $\left( \dfrac{x+\cos x}{\tan x} \right)$ does not exist when $\tan x=0$ i.e. $x=n\pi$. Thus, at the values of $n\pi$ , this function is not differentiable.