AP Calculus AB and BC

Derivative of sin(e^x): Chain Rule

The derivative of sin of e to the x is e to the x times cosine of e to the x. Sine differentiates to cosine with the argument unchanged, and the chain rule multiplies by the derivative of the inner exponential, which is itself.

ddx[sin⁡(ex)]=excos⁡(ex)\frac{d}{dx}\left[\sin\left(e^{x}\right)\right] = e^{x}\cos\left(e^{x}\right)

Sine outside, exponential inside

ddxsin⁡(ex)=cos⁡(ex)⋅ex\frac{d}{dx}\sin\left(e^{x}\right) = \cos\left(e^{x}\right)\cdot e^{x}

The cosine keeps the argument exe^{x}. Writing cos⁡x\cos x there is the error that separates a correct answer from a wrong one.

The oscillation speeds up

As xx grows, exe^{x} races through more and more full turns, so the graph oscillates faster and faster while its amplitude stays at 11. The derivative's exe^{x} factor records exactly that: the slopes grow without bound even though the function does not.

Common mistakes

  • Answering cos⁡(ex)\cos\left(e^{x}\right) and dropping the inner derivative.
  • Answering excos⁡xe^{x}\cos x, evaluating the outer derivative at xx instead of at the inner function.
  • Confusing it with esin⁡xe^{\sin x}, whose derivative is cos⁡x esin⁡x\cos x\,e^{\sin x}.

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 sin(e^x)?

It is excos⁡(ex)e^{x}\cos\left(e^{x}\right).

How is it different from e^(sin x)?

That one differentiates to cos⁡x esin⁡x\cos x\,e^{\sin x}. The layers are swapped, so the answers are different functions.