**Notes:- 1) If y = lnx ; then dy/dx = 1/x**

**2) If y = tanx, then dy/dx = sec^2(x)**

**3) If y = x^n ; where n = constant , then **

**dy/dx = n*{x^(n-1)}**

**3) If 'y' is a function which contains sub-functions, then the last function is differentiated first,then the second last and so on**

Now,

the given question is

`y = tan[ln(ax+b)]`

Thus, **`dy/dx = [sec^2{ln(ax+b)}]*[1/(ax+b)]*a` **

**or, `dy/dx = y' = [a*{sec^2{ln(ax+b)}}]/(ax+b)` **

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now