if ( f(x)=sin(e^{-x}) ), then ( f(x)= ) \n a ( -cos(e^{-x}) ) \n b ( cos(e^{-x})+e^{-x} ) \n c (…

if ( f(x)=sin(e^{-x}) ), then ( f(x)= ) \n a ( -cos(e^{-x}) ) \n b ( cos(e^{-x})+e^{-x} ) \n c ( cos(e^{-x})-e^{-x} ) \n d ( e^{-x}cos(e^{-x}) ) \n e ( -e^{-x}cos(e^{-x}) )

if ( f(x)=sin(e^{-x}) ), then ( f(x)= ) \n a ( -cos(e^{-x}) ) \n b ( cos(e^{-x})+e^{-x} ) \n c ( cos(e^{-x})-e^{-x} ) \n d ( e^{-x}cos(e^{-x}) ) \n e ( -e^{-x}cos(e^{-x}) )

Answer

Explanation:

Step1: Apply the chain rule

The chain rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Let (u = e^{-x}), so (f(x)=\sin(u)). The derivative of (\sin(u)) with respect to (u) is (\cos(u)), and the derivative of (u = e^{-x}) with respect to (x) is (-e^{-x}).

Step2: Calculate the derivative

By the chain rule, (f^\prime(x)=\cos(e^{-x})\cdot(-e^{-x}))

Answer:

E. (-e^{-x}\cos(e^{-x}))