 # Differentiate y = sec(θ) tan(θ). The aim of this problem is to go through the process of differentiation and the use of necessary rules and tables, especially the product rule.

Differentiation is the process in which we calculate the derivative of a given function. There are many rules that ease this process. However, sometimes for some functions, the empirical solution is not that easy and we have to take help from the derivative tables. These tables list the functions and their derivatives as pairs for reference.

In the given question we will have to use the product rule of differentiation. If you are given two functions ( say $u$ and $v$ ) and their derivatives (say u’ and v’) are known, then to find the derivative of their product ( uv ), we use the following product rule:

$\dfrac{ d }{ dx } \bigg ( u v \bigg ) \ = \ u \dfrac{ d }{ dx } \bigg ( v \bigg ) \ + \ v \dfrac{ d }{ dx } \bigg ( u \bigg )$

Let:

$u \ = \ sec(θ) \ \text{ and } \ v \ = \ tan(θ)$

Using derivative tables:

$u’ \ = \ \dfrac{ d }{ dx } \bigg ( sec(θ) \bigg ) \ = \ tan(θ) sec(θ)$

$v’ \ = \ \dfrac{ d }{ dx } \bigg ( tan(θ) \bigg ) \ = \ sec^{ 2 } (θ)$

Given:

$y \ = \ sec(θ) tan(θ)$

$y \ = \ u v$

Differentiating both sides:

$\dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ \dfrac{ d }{ dx } \bigg ( u v \bigg )$

Using the product rule:

$\dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ u \dfrac{ d }{ dx } \bigg ( v \bigg ) \ + \ v \dfrac{ d }{ dx } \bigg ( u \bigg )$

$\dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ u v’ \ + \ v u’$

Substituting values:

$\dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ \bigg ( sec(θ) \bigg ) \bigg ( sec^{ 2 }(θ) \bigg ) \ + \ \bigg ( tan(θ) \bigg ) \bigg ( sec(θ) tan(θ) \bigg )$

$\dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ sec^{ 3 }(θ) \ + \ sec(θ) tan^{ 2 } (θ)$

## Numerical Result

$\dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ sec^{ 3 } (θ) \ + \ sec(θ) tan^{ 2 } (θ)$

## Example

Find the derivative of y = cosec(θ) cot(θ).

$\dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ cosec(θ) \dfrac{ d }{ dx } \bigg ( cot(θ) \bigg ) \ + \ cot(θ) \dfrac{ d }{ dx } \bigg ( cosec(θ) \bigg )$

$\dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ \bigg ( cosec(θ) \bigg ) \bigg ( -cosec^{ 2 }(θ) \bigg ) \ + \ \bigg ( cot(θ) \bigg ) \bigg ( -cosec(θ) cot(θ) \bigg )$

$\dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ – \ cosec^{ 3 }(θ) \ – \ cosec(θ) cot^{ 2 } (θ)$