Solution:

Given That:

[tex]\rm\longrightarrow y = e^{x}-\sin(x)[/tex]

Differentiating both sides with respect to x, we get:

[tex]\rm\longrightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\{ e^{x}-\sin(x)\}[/tex]

We know that:

[tex]\bigstar\:\:\underline{\boxed{\rm\dfrac{d}{dx}(f \pm g) = \dfrac{df}{dx}\pm\dfrac{dg}{dx}}}[/tex]

[tex]\bigstar\:\:\underline{\boxed{\rm\dfrac{d}{dx}e^{x} = e^{x}}}[/tex]

[tex]\bigstar\:\:\underline{\boxed{\rm\dfrac{d}{dx}\sin(x) = \cos(x)}}[/tex]

Using this result, we get:

[tex]\rm\longrightarrow \dfrac{dy}{dx}=\dfrac{d}{dx} e^{x}-\dfrac{d}{dx}\sin(x)[/tex]

[tex]\rm\longrightarrow \dfrac{dy}{dx}=e^{x}-\cos(x)[/tex]

Therefore, option C is the right answer for this problem.

Learn More:

[tex]\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}[/tex]