# Is f(x)=x-e^xsinx increasing or decreasing at x=pi/3?

Apr 16, 2016

Decreasing.

#### Explanation:

Use the sign, positive or negative, of the first derivative of a function to determine whether the function is increasing or decreasing:

• If $f ' \left(\frac{\pi}{3}\right) < 0$, then $f$ is decreasing at $x = \frac{\pi}{3}$.
• If $f ' \left(\frac{\pi}{3}\right) > 0$, then $f$ is increasing at $x = \frac{\pi}{3}$.

So, we first must find $f '$. Note that in order to differentiate the ${e}^{x} \sin x$ term we will have to use the product rule.

After applying the product rule and differentiating the initial $x$ term to get $1$, we see that

$f ' \left(x\right) = 1 - \sin x \frac{d}{\mathrm{dx}} \left({e}^{x}\right) - {e}^{x} \frac{d}{\mathrm{dx}} \left(\sin x\right)$

Recalling that $\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$ and $\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$, this gives

$f ' \left(x\right) = 1 - {e}^{x} \sin x - {e}^{x} \cos x$

So, to determine whether the function is increasing or decreasing, we evaluate the derivative at $x = \frac{\pi}{3}$:

$f ' \left(\frac{\pi}{3}\right) = 1 - {e}^{\pi / 3} \sin \left(\frac{\pi}{3}\right) - {e}^{\pi / 3} \cos \left(\frac{\pi}{3}\right) \approx - 2.8927$

Since this is negative, the function is decreasing at $\frac{\pi}{3}$.

We can check a graph of $f$ (note that $\frac{\pi}{3} \approx 1.0472$):

graph{x-e^xsinx [-9.78, 10.22, -6.75, 3.25]}

At $x \approx 1.0472$, the function is decreasing.