# Find y' and y" if y = ln(sec 2x + tan 2x)

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### 1 Answer

The function `y= ln(sec 2x + tan 2x)`

This can be differentiated using the chain rule.

y' = `(1/(sec 2x + tan 2x))*(2*sec 2x*tan 2x + 2*sec^2 2x)`

=> `(2*sec 2x*tan 2x + 2*sec^2 2x)/(sec 2x + tan 2x)`

=> `2*sec 2x(tan 2x + sec 2x)/(sec 2x + tan 2x)`

=> `2*sec 2x`

y'' = `2*2*sec 2x*tan 2x`

=> `4*sec 2x*tan 2x`

**The derivatives of `y= ln(sec 2x + tan 2x)` are `y' = 2*sec 2x ` and `y''= 4*sec 2x*tan 2x` **