What is the second derivative of # e^x(cosx-sinx) #?

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Answer 1 · 2017-04-03T13:05:33

And doing the same again we get:

# (d^2)/(dx^2) e^x(cosx-sinx)= -2 e^x(cosx + sinx) #

Before tackling the problem I will derive two useful derivative results using the product rule:

# d/dx e^xsinx = e^xd/dxsinx + d/dxe^xsinx #
# " " = e^xcosx + e^xsinx #

And:

# d/dx e^xcosx = e^xd/dxcosx + d/dxe^xcosx #
# " " = -e^xsinx + e^xcosx #
# " " = e^xcosx -e^xsinx #

Let:

# y = e^x(cosx-sinx) #
# \ \ = e^xcosx-e^xsinx #

Then using the above derivative results we have:

# dy/dx = (e^xcosx -e^xsinx) - (e^xcosx + e^xsinx) #
# " " = e^xcosx -e^xsinx - e^xcosx - e^xsinx #
# " " = -2 e^xsinx #

And doing the same again we get:

# (d^2y)/(dx^2) = -2 (e^xcosx + e^xsinx) #
# " " = -2 e^x(cosx + sinx) #