why is the derivative of $$(Ax+B)e^x$$
$$(Ax+A+B)e^x$$?
I understand the Ax+A part, but why didn't the B disappear?

btw it's an ODE problem. $$y''-4y=xe^x+cos2x$$

Question

why is the derivative of $$(Ax+B)e^x$$
$$(Ax+A+B)e^x$$?
I understand the Ax+A part, but why didn't the B disappear?

btw it's an ODE problem. $$y''-4y=xe^x+cos2x$$

turingtest · Answer

$$y=(Ax+B)e^x$$$$y'=(Ax+B)e^x+Ae^x=(Ax+A+B)e^x$$

anonymous · Answer

how odd, wolfram alpha won't show me the steps unless I sign up with them.

anonymous · Answer

why doesn't the B disappear? Isn't it a constant?

phi · Answer

derivative of B * e^x  is B* d/dx (e^x) = B* e^x

turingtest · Answer

By the product rule the B disappears from one Ax+B term, and in the other term you differentiate e^x
$$u=Ax+B$$$$u'=A$$$$v=e^x$$$$v'=e^x$$$$uv'+vu'=(Ax+B)e^x+Ae^x$$factor out $e^x$ and you get$$(Ax+A+B)e^x$$

anonymous · Answer

awesome thanks!!!!