y = tan xy=tanx
dy/dx = sec^2 xdydx=sec2x
This is a common identity which many people memorize, along with the derivatives of sin xsinx and cos xcosx because they appear so frequently.
It can be proven easily using two well-known trig identities and the quotient rule.
Process:
First recall from pre-calculus that tan theta = sin theta / cos thetatanθ=sinθcosθ.
So, we can rewrite y = tan xy=tanx equivalently as:
y = sin x / cos xy=sinxcosx
Now, we differentiate, and apply the quotient rule:
dy/dx = (d/dx[sinx]*cosx - d/dx[cosx]*sinx)/(cos x)^2dydx=ddx[sinx]⋅cosx−ddx[cosx]⋅sinx(cosx)2
We know that the derivative of sinxsinx is cosxcosx, and that the derivative of cosxcosx is -sinx−sinx. So, upon simplifying the above equation, we arrive at:
dy/dx = (cos^2 x + sin^2 x)/(cos^2 x)dydx=cos2x+sin2xcos2x
It should be clear that the numerator can be simplified using another trig identity.
Recall sin^2 theta + cos^2 theta = 1sin2θ+cos2θ=1. (or the Pythagorean identity)
Substitution yields:
dy/dx = 1/(cos^2 x)dydx=1cos2x
which is equivalent to:
dy/dx = sec^2 xdydx=sec2x