can someone help me with this? find the first 5 derivatives of the function f(x)=e^x cosx, then find the 41st derivative. If possible, please show the work, thanks :)

Question

precal · Answer

Use the product rule.  Looks like your teacher wanted to keep you busy.....

jamesj · Answer

Cute question.  What are the first four derivatives?

We can detect a pattern; the idea obviously is not to differentiate 41 times.

jamesj · Answer

What's f'(x)?

precal · Answer

You still have to use the product rule..

anonymous · Answer

But is there a short cut to get to the 41st derivative? I mean, my professor can't seriously expect me to do this :(

precal · Answer

Do the first four .........

jamesj · Answer

Yes there is.  That's why s/he asked you to do the first 4 first.  So what are they?

anonymous · Answer

To be honest, I'm not even completely sure how to do those either. I have an exam tmrw, so I'm trying to understand as much as I can...

jamesj · Answer

If f(x) = e^x sin x, what is f'(x), just the first derivative?

jamesj · Answer

product rule

jamesj · Answer

sorry, let's stay with your function, f(x) = e^x . cos x

anonymous · Answer

ok,so derivative of cosx is -sinx, right?

jamesj · Answer

Yep, and I hope you've got that engraved in your brain for tomorrow!

anonymous · Answer

Uh, actually no. Sad to say I had to look that up, lol...

precal · Answer

First derivative is 
$$e ^{x}cosx-sinx(e ^{x})$$
Factor out $$e ^{x}$$
$$e ^{x}(cosx-sinx)$$
You need to do your homework, because you can't wing calculus..

anonymous · Answer

I know, I don't know how I even got into this class. Math is my worst subject :(

precal · Answer

sorry, talk to your teacher......  :(     I wish I could tell you more....

anonymous · Answer

Yea, I wish I could, but he just makes everything sound more complicated. He gives us the question and the answer, but nothing in between, so idk how to do the work.

jamesj · Answer

f'(x) = e^x (cos x - sin x) = f(x) -  e^x sin x

So
f''(x) = f'(x) - e^x (sin x + cos x)
       = e^x ( cos x - sin x - sin x - cos x)
       = - 2 e^x sin x

Now 
f'''(x) = - 2 e^x (sin x + cos x)
f''''(x) = -2 e^x (sin x + cos x + cos x - sin x)
         = -4 e^x cos x
         = -4 f(x)

jamesj · Answer

Thus
$$f^{(40)}(x) = (-4)^{10} f(x) = 2^{20} f(x)$$
and
$$f^{(41)}(x) = 2^{20} f'(x) = 2^{20}e^x(\cos x - \sin x)$$

anonymous · Answer

but i don't understand how for the 4th derivative, why did the -2 become -4?

anonymous · Answer

thats weird, looking back at the answer my professor gave me, he wrote that the 41st derivative is (-4^10)(-e^sinx+e^xcosx)

jamesj · Answer

What I have written is exactly equal.

anonymous · Answer

Oh, really? Well I really liked how you showed your steps. Are you a teacher or something?

jamesj · Answer

I was

anonymous · Answer

where did you teach?