What is the derivative of e^(-x)?

Question

I have to integrate e^(-x) * cos(2x) dx and I'm using integration by parts. u=e^(-x) dv=cos(2x) dx. Is that wrong and if not, how do you do you take the derivative of that exp. function.

Answer 1

(d/dx)e^-x = - e^-x

Answer 2

The derivative of e^(-x) is -e^(-x). That's the correct method, doing parts gives an integral of e^(-x) * sin(2x), so use parts again with dv = sin(2x), leaving an e^(-x) * cos(2x) integral, then take this to the other side of the equals, rearrange to make the integral the subject, leaving the answer as e^(-x)/3 * (2sin(2x) - cos(2x))

Let me know if you need further explanation

Answer 3

u = e^(-x)

du/dx = -e^(-x)

I think it is easier to have them the other way around. Integrating exponentials is easier than integrating trigonometric functions.

My old teacher gave us the LIATE rule, a mnemonic which should be u and which should be dv. The higher up it is, the better it is to take it as the u term.

Logarithms, Inverse trigonometric, Algebraic, Trigonometric, Exponential.

Try to keep exponentials as the dv term.

Answer 4

The derivative of e ^ x is e ^ x.

But remember, since there is a constant next to the 'x', the negative turns into a negative one and gets brought down.

So y'  =  -e ^ x. And that's your answer.

Answer 5

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Answer 6

e^(-x) (=e^u)= -e^(-x)

normally a^x=(a^x)*ln(a)