Picknell14
Answered

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What is the derivative of y=ln(sec(x)+tan(x))y=ln(sec(x)+tan(x))?

Sagot :

kittymaniac
The derivative of ln(u) is 1/u * the derivative of u.

[tex]y = ln(sec(x)+tan(x)) \\ \\ y' = \frac{1}{sec(x)+tan(x)} * [tan(x)sec(x) + sec^2(x)] \\ \\ y' = \frac{sec(x)(sec(x) + tan(x))}{sec(x)+tan(x)} \\ \\ y' = sec(x) + C[/tex]

I hope this helps! :)
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