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How do you find the derivative of #2e^-x#?

1 Answer

Oct 31, 2016

#(dy)/(dx)=-2e^-x#

Explanation:

Recall that #d/dxe^x=e^x#

Using chain rule, #(dy)/(dx)=(dy)/(du)*(du)/(dx)#,

Let #u=-x#

#(dy)/(du)=d/(du)2e^u=2e^u=2e^-x#

#(du)/(dx)=d/(dx)-x=-1#

#:.(dy)/(dx)=-1*2e^-x=-2e^-x#

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