AP Calculus AB and BC

Derivative of e^(tan x): Chain Rule

The derivative of e to the tan x is secant squared x times e to the tan x. The exponential differentiates to itself with the exponent unchanged, then the chain rule multiplies by the derivative of the exponent, which is secant squared x.

ddx[etan⁡x]=sec⁡2x etan⁡x\frac{d}{dx}\left[e^{\tan x}\right] = \sec^{2}x\,e^{\tan x}

Exponential outside, tangent inside

ddxeu=euu′  ⟹  ddxetan⁡x=etan⁡xsec⁡2x\frac{d}{dx}e^{u} = e^{u}u' \implies \frac{d}{dx}e^{\tan x} = e^{\tan x}\sec^{2}x

The exponent stays tan⁡x\tan x in the answer. Only the multiplier is new.

Always positive, and steep

Both factors are positive wherever they exist, so the function is increasing on every interval between the tangent asymptotes. Near x=π2x = \frac{\pi}{2} the exponent runs to infinity and the function grows extremely fast.

Common mistakes

  • Answering esec⁡2xe^{\sec^{2}x}, putting the derivative into the exponent.
  • Answering etan⁡xe^{\tan x} alone, dropping the chain rule factor.
  • Using sec⁡xtan⁡x\sec x\tan x, which is the derivative of secant rather than of tangent.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of e^(tan x)?

It is sec⁡2x etan⁡x\sec^{2}x\,e^{\tan x}.

Does the exponent change?

No. It stays tan⁡x\tan x; the chain rule adds a factor in front.