91Ó°ÊÓ

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

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,

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

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

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

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

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

Since $f\text{'}\left(x\right)>0,$ thus therole="math" localid="1649845993278" $f\text{'}\mathrm{is}\mathrm{positive}\mathrm{on}\mathrm{the}\mathrm{interval}\left(-∞,\frac{\mathrm{π}}{4}\right).$

Now, the interval$\left(\frac{\mathrm{π}}{4},∞\right),$

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

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