Given ,
y=sin(ex)y=sin(e^x)y=sin(ex)
Differentiate y With respect to x using chain rule
→y′=cos(ex)×d(ex)dx\to y'=cos(e^x)\times\dfrac{d(e^x)}{dx}→y′=cos(ex)×dxd(ex)
→y′=cos(ex)×ex\to y'=cos(e^x)\times e^x→y′=cos(ex)×ex
→y′=excos(ex)\to y'=e^xcos(e^x)→y′=excos(ex)
So option (A) is correct.
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