Question

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

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