`y = tan[ln(ax + b)]` Differentiate the function.

Textbook Question

Chapter 3, 3.6 - Problem 17 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385 | (Level 1) Assistant Educator

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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)`

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