Proof that d/dx tax = sec^2(x)

Could you demonstrate this proof to me?

Unlike your last post, which dealt with the very basic functions \sin and \cos, we can express \tan as \displaystyle \frac{\sin}{\cos}, so we have more to work with.

\frac{d}{dx} \tan x = \frac{d}{dx}\frac{\sin x}{\cos x} = \underbrace{\frac{\sin^2 x + \cos^2 x}{\cos^2 x}}_{\text{using the quotient rule}} = \frac{1}{\cos^2 x} = \sec^2 x.

Let me know if you have any questions.

Thanks! That makes sense

No problem