How do you find the derivative of y=tan(x)y=tan(x) ?

1 Answer
Aug 1, 2014

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]cosxddx[cosx]sinx(cosx)2

We know that the derivative of sinxsinx is cosxcosx, and that the derivative of cosxcosx is -sinxsinx. 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