What is the second derivative of e^x(cosx-sinx) ?
1 Answer
Apr 3, 2017
And doing the same again we get:
(d^2)/(dx^2) e^x(cosx-sinx)= -2 e^x(cosx + sinx)
Explanation:
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)