+4 When in terms of derivative, what does secant of x, sec (x), equal to? I am still confused as to how to come up with the answer :)
+118654 When in terms of derivative, what does secant of x, sec (x), equal to? I am still confused as to how to come up with the answer :)
\(\frac{d}{dx}\;sec(x)\\ =\frac{d}{dx}\;(cos(x))^{-1}\\ =-(cos(x))^{-2}\times -sin(x)\\ =\frac{sinx}{cos^2(x)}\\ =tan(x)sec(x)\)
+118654 More examination:
I have used the chain rule to answer your question. This is a more formal setting out.
It is a good idea to understand what is really happening.
I had a good teacher at school but she did everything with 'short cuts' and I am still learning some of the formality that lies behind those short cuts.
I cannot answer some more complex differentials unless I understand the principals. So I find some things more difficult than I should.
When in terms of derivative, what does secant of x, sec (x), equal to? I am still confused as to how to come up with the answer :)
NOTE: g is a funtion of x so I could have called it g(x)
\(let\;\; g=cos(x)\\ y=g^{-1}\\ \frac{dy}{dx}=\frac{dy}{dg}\times \frac{dg}{dx}\\ \text{The gs cancel out}\\ \frac{dy}{dx}=-g^{-2}\times-sin(x)\\ \frac{dy}{dx}=-(cos(x))^{-2}\times-sin(x)\\ etc\)
