**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:

\[ \dfrac{ d }{ dx } \bigg ( u v \bigg ) \ = \ u \dfrac{ d }{ dx } \bigg ( v \bigg )Â \ + \ v \dfrac{ d }{ dx } \bigg ( u \bigg ) \]

## Expert Answer

**Let:**

\[ u \ = \ sec(Î¸) \ \text{ and } \ v \ = \ tan(Î¸) \]

**Using derivative tables:**

\[ u’ \ = \ \dfrac{ d }{ dx } \bigg ( sec(Î¸) \bigg ) \ = \ tan(Î¸) sec(Î¸)\]

\[ v’ \ = \ \dfrac{ d }{ dx } \bigg ( tan(Î¸) \bigg ) \ = \ sec^{ 2 } (Î¸)\]

**Given:**

\[ y \ = \ sec(Î¸) tan(Î¸) \]

\[ y \ = \ u v \]

**Differentiating both sides:**

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

**Using the product rule:**

\[ \dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ u \dfrac{ d }{ dx } \bigg ( v \bigg )Â \ + \ v \dfrac{ d }{ dx } \bigg ( u \bigg ) \]

\[ \dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ u v’Â \ + \ v u’ \]

**Substituting values:**

\[ \dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ \bigg ( sec(Î¸) \bigg ) \bigg ( sec^{ 2 }(Î¸) \bigg ) Â \ + \ \bigg ( tan(Î¸) \bigg ) \bigg ( sec(Î¸) tan(Î¸) \bigg ) \]

\[ \dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ sec^{ 3 }(Î¸) \ + \ sec(Î¸) tan^{ 2 } (Î¸) \]

## Numerical Result

\[ \dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ sec^{ 3 } (Î¸) \ + \ sec(Î¸) tan^{ 2 } (Î¸) \]

## Example

Find the **derivative of y = cosec(Î¸) cot(Î¸).**

\[ \dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ cosec(Î¸) \dfrac{ d }{ dx } \bigg ( cot(Î¸) \bigg )Â \ + \ cot(Î¸) \dfrac{ d }{ dx } \bigg ( cosec(Î¸) \bigg ) \]

\[ \dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ \bigg ( cosec(Î¸) \bigg ) \bigg ( -cosec^{ 2 }(Î¸) \bigg ) Â \ + \ \bigg ( cot(Î¸) \bigg ) \bigg ( -cosec(Î¸) cot(Î¸) \bigg ) \]

\[ \dfrac{ d }{ dx } \bigg ( y \bigg ) \ = \ – \ cosec^{ 3 }(Î¸) \ – \ cosec(Î¸) cot^{ 2 } (Î¸) \]