Question

How do you find the derivative of $ \dfrac{\sec x+\text{cosec }x}{\text{cosec }x} $

Answer 1

Hint: Derivative of a function is the rate of change in the function with respect to the independent variable (x). First try to simplify the given function. Then use the properties of differentiation and differentiate the function.

Complete step-by-step solution:
Function is simply a quantity or a variable that depends on another quantity or variable. In the given equation y is a function that is depending on the variable x.
Let us now understand what is meant by the derivative of a function.
Derivative of a function is the rate of change in the function with respect to the independent variable (x). In other words, it is the rate at which the function y changes when the value of x is changed.
The derivative of y with respect to x is denoted as $ \dfrac{d}{dx}(y) $ , where $ y=f(x) $ .
We also called this as differentiation of x with respect to x.
i.e. $ \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \dfrac{\sec x+\text{cosec }x}{\text{cosec }x} \right) $ …. (i)
Before differentiating let us first simplify the given equation.
We can write the equation as $ \dfrac{\sec x+\text{cosec }x}{\text{cosec }x}=\dfrac{\sec x}{\text{cosec }x}+1 $ .
And we know that $ \dfrac{\sec x}{\text{cosec }x}=\tan x $
Therefore, we can write that $ \dfrac{\sec x+\text{cosec }x}{\text{cosec }x}=\tan x+1 $
With this, equation (i) changes to $ \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \tan x+1 \right) $
Since differentiation has associative property, $ \dfrac{d}{dx}\left( \tan x+1 \right)=\dfrac{d}{dx}(\tan x)+\dfrac{d}{dx}(1) $
But we know that differentiation of a constant term is zero. Therefore, $ \dfrac{d}{dx}(1) $ .
And $ \dfrac{d}{dx}(\tan x)={{\sec }^{2}}x $ .
Therefore, we get that $ \dfrac{dy}{dx}=\dfrac{d}{dx}\left( \tan x+1 \right)={{\sec }^{2}}x $ .
Hence, we found the first derivative of the given function.

Note: Whenever the question is asked to find the derivative of function involving trigonometric functions, first simply the complex function in the terms of sine, cosine and tangent function as we know the derivatives of these functions. We can find the derivative of the given function using quotient rule but it will be a longer process.