Differentiate e^(ax)cos(2x) and find stationary points exactly

Expert intro: This one rewards tidy differentiation and a careful algebraic clean-up afterward. The exponential factor never vanishes, so the stationary points come from the trig expression only.

Detailed walkthrough

Given

$y=e^{\frac{2}{3}\sqrt{3}x}\cos 2x,\qquad 0<x<\pi,$

let

$a=\frac{2\sqrt{3}}{3}.$

Then

$y=e^{ax}\cos 2x.$

(a) Differentiate

Use the product rule:

$\frac{dy}{dx}=\frac{d}{dx}(e^{ax})\cos 2x+e^{ax}\frac{d}{dx}(\cos 2x).$

So

$\frac{dy}{dx}=ae^{ax}\cos 2x-2e^{ax}\sin 2x.$

Factor out $e^{ax}$:

$\frac{dy}{dx}=e^{ax}(a\cos 2x-2\sin 2x).$

Substitute back $a=\frac{2\sqrt{3}}{3}$:

$\boxed{\frac{dy}{dx}=e^{\frac{2}{3}\sqrt{3}x}\left(\frac{2\sqrt{3}}{3}\cos 2x-2\sin 2x\right)}.$

(b) Find stationary points exactly

Stationary points occur when $\frac{dy}{dx}=0$.

Since $e^{\frac{2}{3}\sqrt{3}x}>0$ for all $x$, we only need

$\frac{2\sqrt{3}}{3}\cos 2x-2\sin 2x=0.$

Divide by 2:

$\frac{\sqrt{3}}{3}\cos 2x-\sin 2x=0.$

Rearrange:

$\sin 2x=\frac{\sqrt{3}}{3}\cos 2x.$

If $\cos 2x\ne 0$, divide through by $\cos 2x$:

$\tan 2x=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}.$

So

$2x=\frac{\pi}{6}+k\pi.$

Because $0<x<\pi$, we have $0<2x<2\pi$, so the valid values are

$2x=\frac{\pi}{6},\ \frac{7\pi}{6}.$

Hence

$x=\frac{\pi}{12},\ \frac{7\pi}{12}.$

Now find the corresponding $y$-values.

At $x=\frac{\pi}{12}$

= e^{\frac{\pi\sqrt{3}}{18}}\cos\left(\frac{\pi}{6}\right) = e^{\frac{\pi\sqrt{3}}{18}}\cdot \frac{\sqrt{3}}{2}.$$ #### At $x=\frac{7\pi}{12}$ $$y=e^{\frac{2\sqrt{3}}{3}\cdot \frac{7\pi}{12}}\cos\left(\frac{7\pi}{6}\right) = e^{\frac{7\pi\sqrt{3}}{18}}\cdot\left(-\frac{\sqrt{3}}{2}\right).$$ So the stationary points are $$\boxed{\left(\frac{\pi}{12},\frac{\sqrt{3}}{2}e^{\frac{\pi\sqrt{3}}{18}}\right)}$$ and $$\boxed{\left(\frac{7\pi}{12},-\frac{\sqrt{3}}{2}e^{\frac{7\pi\sqrt{3}}{18}}\right)}.$$ ### 💡 Pitfall guide The most common error is to lose the minus sign when differentiating $\cos 2x$; it must become $-2\sin 2x$. Another trap is to stop at $\tan 2x=1/\sqrt{3}$ and forget the interval $0<2x<2\pi$, which gives two solutions, not one. Also, do not try to set the exponential factor equal to zero — it never is. ### 🔄 Real-world variant If the cosine term were $\cos nx$ instead of $\cos 2x$, the derivative would become $e^{ax}(a\cos nx-n\sin nx)$. The stationary-point condition would change to $\tan nx=a/n$, so the same strategy works, but the exact angles depend on $n$. ### 🔍 Related terms product rule, stationary points, trigonometric equation