Critical points of $f(x)=e^{-x} sinx$ on $[0, 2 \pi]$

Question

luigi0210 · Answer

I got $$\large f'(x)=e^{-x}(cosx-sinx)$$
$$\large 0=e^{-x}(cosx-sinx)$$?

nincompoop · Answer

how did you differentiate e^-x * sin x?

luigi0210 · Answer

$$\LARGE f'(x)=e^{-x} cosx-sinx e^{-x}$$
$$\LARGE =e^{-x}(cosx-sinx)$$

luigi0210 · Answer

$$\large f'(x)=(e^{-x})(sinx)'-(sinx)(e^{-x})'$$

nincompoop · Answer

cool
then obtain the points between the interval given at which the value or tangent is zero

luigi0210 · Answer

Well I know you can't do $\large e^{-x}=0$ but $cosx-sinx=0$?

nincompoop · Answer

e^-x by itself, no solution exist. meaning that it will never be zero

luigi0210 · Answer

$$\large cosx=sinx$$?

nincompoop · Answer

yes?

luigi0210 · Answer

OH, now I see what you mean.. took me awhile. Thanks nin!

nincompoop · Answer

trig can be confusing and isn't very intuitive to find critical points 
you did the right thing by factoring, because that is when you can easily determine zeros