Question

The derivative of \(\sec (x) \equiv \frac{1}{\cos (x)}\) is?

Answer 1

\[\frac{d}{dx}\sec(x)=\sec(x)\tan(x)\]

Answer 2

no do it step by step according to this method:-
\(f (x)\) has a derivative \(f' (x)\) and \(f (x) \neq 0\) then

\(\left( \frac{1}{f} \right)' (x) = \frac{- f' (x)}{\left[ f (x) \right]^2}\)

Answer 3

\(\frac{? (x)}{\cos^2 (x)}\)

Answer 4

using the quotient rule? alright.
\[u := 1\]
\[v := \cos(x)\]
\[\frac{d}{dx}\frac{u}{v}=\frac{\frac{du}{dx}v - \frac{dv}{dx}u}{v^2}\]
\[\frac{d}{dx}\sec(x)=\frac{d}{dx}\frac{1}{\cos(x)}=\frac{-\sin(x)}{\cos^2(x)}\]

Answer 5

got it

Answer 6

there's a mistake there, it should be \[\frac{\sin(x)}{\cos^2(x)}\]

Answer 7

ya i know