Find derivative of ln[sec x]
Ok... I know the derivative of ln x = 1/x ... But when it has a function inside of it, what do I do? do I still do 1/sec x then use the chain rule?

Question

Find derivative of ln[sec x]
Ok... I know the derivative of ln x = 1/x ... But when it has a function inside of it, what do I do? do I still do 1/sec x then use the chain rule?

johnweldon1993 · Answer

Yeah that looks right...

So we would have

$$\large \frac{1}{\sec(x)} \times \sec(x)\tan(x)$$ right?

anonymous · Answer

Yeah... Still gotta remember those damn trig derivatives and the antiderivatives, and then derive them for the AP test! lol...

anonymous · Answer

Thanks for the answer, anything else before I give out the best response? like any exceptions or whatever?

johnweldon1993 · Answer

Nope that's about it...but what do you get when you simplify that?

kc_kennylau · Answer

You only have to memorize $\dfrac d{dx}\sin x=\cos x$ and $\dfrac d{dx}\cos x=-\sin x$ and you can derive all of them using quotient rule.

anonymous · Answer

wait... just realized that isn't one of the answers... so maybe if I convert it in terms of cos and sin...
since sec x is just 1/cos x, it would be
$$(\cos x ) + \frac{ 1 }{ \cos x} * \tan x$$
so just tan x, which is sinx / cox

johnweldon1993 · Answer

Right...the simplification would be tan(x)

johnweldon1993 · Answer

$$\large \frac{1}{sec(x)} \times \sec(x)\tan(x) = \frac{\sec(x)\tan(x)}{\sec(x)} = \tan(x)$$

kc_kennylau · Answer

It's not plus; it's multiply @Psychedelic

anonymous · Answer

got that, thanks!