The derivative of secant is ddx[sec⁡(x)]=sec⁡(x)tan⁡(x)\frac{d}{dx}[\sec(x)]=\sec(x)\tan(x)dxd​[sec(x)]=sec(x)tan(x). This is the standard trig derivative rule used in calculus.

Quick derivation

Since sec⁡(x)=1cos⁡(x)\sec(x)=\frac{1}{\cos(x)}sec(x)=cos(x)1​, differentiating gives:

ddx(sec⁡(x))=ddx(1cos⁡(x))=sin⁡(x)cos⁡2(x)=sec⁡(x)tan⁡(x)\frac{d}{dx}\left(\sec(x)\right) = \frac{d}{dx}\left(\frac{1}{\cos(x)}\right) = \frac{\sin(x)}{\cos^2(x)} = \sec(x)\tan(x)dxd​(sec(x))=dxd​(cos(x)1​)=cos2(x)sin(x)​=sec(x)tan(x)

That matches the standard result shown in multiple calculus references.

If you meant a chain rule version

For sec⁡(g(x))\sec(g(x))sec(g(x)), the derivative is:

ddx[sec⁡(g(x))]=sec⁡(g(x))tan⁡(g(x))g′(x)\frac{d}{dx}[\sec(g(x))]=\sec(g(x))\tan(g(x))g'(x)dxd​[sec(g(x))]=sec(g(x))tan(g(x))g′(x)

This follows the same rule with an added chain rule factor.