Q. 3 - Calculus

Question

Sign analyses for derivatives: For each function f that follows, find the derivative $f\text{'}.$ Then determine the intervals on which the derivative $f\text{'}$  is positive and the intervals on which the derivative $f\text{'}$ is negative. Record your answers on a sign chart for $f\text{'},$ with tick-marks only at the x-values where$f\text{'}$ is zero or undefined.

$f\left(x\right)=\frac{\sin x}{e^x}$

Step-by-Step Solution

Answer

The derivative of the given function is positive on the interval $\left(-\infty ,\frac{\mathrm{\pi }}{4}\right)$ and negative on the interval $\left(\frac{\mathrm{\pi }}{4},\infty \right).$ The derivative of the function is zero at $x=\frac{\mathrm{\pi }}{4}.$

Step 1. Given Information.

The given function is $f\left(x\right)=\frac{\sin x}{e^x}.$

Step 2. Find the derivative.

To find the derivative of the given function, we will use the quotient rule of differentiation.

So,

$$\begin{gathered} f\left(x\right)=\frac{\sin x}{e^x} \\ f\text{'}\left(x\right)=\frac{e^x\left(\cos x\right)-\sin x\left(e^x\right)}{{\left(e^x\right)}^2} \\ f\text{'}\left(x\right)=\frac{e^x(\cos x-\sin x)}{{\left(e^x\right)}^2} \\ f\text{'}\left(x\right)=\frac{\cos x-\sin x}{e^x} \end{gathered}$$

Now, let's find the critical point by putting the above equation to zero,

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\cos x-\sin x}{e^x} \\ 0=\frac{\cos x-\sin x}{e^x} \\ 0=\cos x-\sin x \\ \tan x=1 \\ x=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Thus, the critical point is $x=\frac{\mathrm{\pi }}{4}.$ 

Step 3. Determine the intervals.

The intervals we get by the critical points are $\left(-\infty ,\frac{\mathrm{\pi }}{4}\right),\left(\frac{\mathrm{\pi }}{4},\infty \right).$

Now, let's take the interval $\left(-\infty ,\frac{\mathrm{\pi }}{4}\right),$ to determine where the derivative of the function is positive or negative.

$$\begin{gathered} \mathrm{Let} x=-\frac{\mathrm{\pi }}{2} \\ f\text{'}(-\frac{\mathrm{\pi }}{2})=\frac{\cos \left(-\frac{\mathrm{\pi }}{2}\right)-\sin \left(-\frac{\mathrm{\pi }}{2}\right)}{e^{-\frac{\mathrm{\pi }}{2}}} \\ f\text{'}(-\frac{\mathrm{\pi }}{2})=4.8 \end{gathered}$$

Since $f\text{'}\left(x\right)>0,$ thus the  $f\text{'} \mathrm{is} \mathrm{positive} \mathrm{on} \mathrm{the} \mathrm{interval} \left(-\infty ,\frac{\mathrm{\pi }}{4}\right).$

Now, the interval $\left(\frac{\mathrm{\pi }}{4},\infty \right),$

$$\begin{gathered} \mathrm{Let} x=\frac{\mathrm{\pi }}{2} \\ f\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=\frac{\cos \left(\frac{\mathrm{\pi }}{2}\right)-\sin \left(\frac{\mathrm{\pi }}{2}\right)}{e^{\frac{\mathrm{\pi }}{2}}} \\ f\text{'}\left(\frac{\mathrm{\pi }}{2}\right)=-0.2 \end{gathered}$$

Since $f\text{'}\left(x\right)<0,$  thus the $f\text{'} \mathrm{is} \mathrm{negative} \mathrm{on} \mathrm{the} \mathrm{interval} \left(\frac{\mathrm{\pi }}{4},\infty \right).$