To find [tex]\(\frac{dy}{dx}\)[/tex] given [tex]\(x = e^t + \sin(t)\)[/tex] and [tex]\(y = e^t - \cos(t)\)[/tex], we can proceed as follows:
1. Compute [tex]\( \frac{dy}{dt} \)[/tex]:
Given [tex]\( y = e^t - \cos(t) \)[/tex], we differentiate [tex]\( y \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[
\frac{dy}{dt} = \frac{d}{dt} (e^t - \cos(t))
\][/tex]
Using the differentiation rules for [tex]\( e^t \)[/tex] and [tex]\(\cos(t)\)[/tex], we get:
[tex]\[
\frac{dy}{dt} = e^t - (-\sin(t)) = e^t + \sin(t)
\][/tex]
2. Compute [tex]\( \frac{dx}{dt} \)[/tex]:
Given [tex]\( x = e^t + \sin(t) \)[/tex], we differentiate [tex]\( x \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[
\frac{dx}{dt} = \frac{d}{dt} (e^t + \sin(t))
\][/tex]
Using the differentiation rules for [tex]\( e^t \)[/tex] and [tex]\(\sin(t)\)[/tex], we get:
[tex]\[
\frac{dx}{dt} = e^t + \cos(t)
\][/tex]
3. Form the ratio [tex]\( \frac{dy}{dx} \)[/tex]:
We use the chain rule to get [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[
\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{e^t + \sin(t)}{e^t + \cos(t)}
\][/tex]
Therefore, the simplified form of [tex]\( \frac{dy}{dx} \)[/tex] is:
[tex]\[
\frac{e^t + \sin(t)}{e^t + \cos(t)}
\][/tex]
Comparing this with the options provided:
[tex]\[
\boxed{\frac{e^t + \sin(t)}{e^t + \cos(t)}}
\][/tex]
Hence, the correct answer is indeed:
[tex]\[
\boxed{(B) \frac{e^t + \sin(t)}{e^t + \cos(t)}}
\][/tex]
