Question

How do you find the derivative of \[y=\tan x+\cot x\]?

Answer 1

Hint: We are given an expression having trigonometric functions and we are asked to find its derivative. We will take each of the terms and find its derivative. We know that the derivative of \[\tan x\] is \[{{\sec }^{2}}x\] and also the derivative of \[\cot x\] is \[-{{\csc }^{2}}x\]. We will apply this accordingly and find the derivative of the given expression.

Complete step by step solution:
According to the given question, we are given an expression and we have to find the derivative of that expression. We can see that the expression has the trigonometric functions.
The expression we have is,
\[y=\tan x+\cot x\]----(1)
Taking the derivative on both sides of equation (1), we will get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}(\tan x)+\dfrac{d}{dx}(\cot x)\]----(2)
Now, we will write the respective derivative of the trigonometric function. We know that, the derivative of \[\tan x\] is \[{{\sec }^{2}}x\] and also the derivative of \[\cot x\] is \[-{{\csc }^{2}}x\].
So, we will substitute these values in the equation (2),
\[\Rightarrow \dfrac{dy}{dx}={{\sec }^{2}}x+(-{{\csc }^{2}}x)\]----(3)
We will now open up the brackets in the equation (3), and write the equation taking in consideration the negative sign. The new expression we now have is,
\[\Rightarrow \dfrac{dy}{dx}={{\sec }^{2}}x-{{\csc }^{2}}x\]---(4)
Therefore, the derivative of the given expression is \[\dfrac{dy}{dx}={{\sec }^{2}}x-{{\csc }^{2}}x\].

Note: The derivative of the trigonometric functions should be written correctly and with correct signs (either positive or negative). Also, the derivative of the given expression should be computed step wise to prevent skipping or overlooking any terms. The derivative of all the trigonometric functions should be known beforehand and also it should not get mixed up while writing the answers.