Could somebody give me a hand with this question? What is the derivative of
y = sec θ tan θ

Question

Could somebody give me a hand with this question?  What is the derivative of 
y = sec θ tan θ

anonymous · Answer

you can use the product rule, or you can use quotient rule after rewriting this as 
$$y=\frac{\sin(x)}{\cos^2(x)}$$

anonymous · Answer

probably easier to use 
$$(fg)'=f'g+g'f$$
with $f(x)=\sec(x), f'(x)=\sec(x)\tan(x), g(x)=\tan(x),g'(x)=\sec^2(x)$

anonymous · Answer

I tired using the product rule and got sec*sec^2+tan*sec*csc

anonymous · Answer

I would assume that is not correct =).

anonymous · Answer

there is no cosecant in it

anonymous · Answer

derivative of secant is secant tangent, not secant cosecant
that is your only mistake

anonymous · Answer

If I rewrote the question in terms of division, wouldn't I have to use the chain rule?