Derivative of e^(cos x): Answer, Proof, Mistakes

How to differentiate e^(cos x)

The outer function is $e^{u}$ and the inner is $u = \cos x$. Differentiating the outer leaves $e^{\cos x}$ untouched, and the chain rule then multiplies by $u' = -\sin x$.

Reversing the work confirms it: the substitution $u = \cos x$ gives $\int -\sin x\,e^{\cos x}\,dx = e^{\cos x} + C$, so the derivative and the antiderivative line up.

What the derivative says about the graph

$e^{\cos x}$ is positive for every $x$, so the sign of $f'$ is decided entirely by $-\sin x$. The function rises where $\sin x<0$ and falls where $\sin x>0$.

Critical points land at multiples of $\pi$, where $\sin x = 0$. The maxima sit at $x = 0,\ \pm 2\pi,\ \dots$ with height $e$, and the minima at $x = \pm\pi,\ \pm 3\pi,\ \dots$ with height $\frac{1}{e}$.

The whole picture repeats with period $2\pi$, a bounded wave rather than the runaway growth of $e^{x}$.

Common mistakes with the derivative of e^(cos x)

• Answering $e^{\cos x}$ by itself. The exponential returns unchanged only when the exponent is $x$ alone.
• Answering $e^{-\sin x}$, differentiating the exponent in place instead of multiplying by it.
• Dropping the minus sign and writing $\sin x\,e^{\cos x}$, which reverses every increasing and decreasing interval.
• Confusing $e^{\cos x}$ with the product $e^{x}\cos x$, which needs the product rule and gives $e^{x}(\cos x-\sin x)$.