Question

what is the nth differential coefficient of e^(2x) cos4x ?​

Answer 1

Answer:

dndxn(uv)=∑k=0n(nk)dkudxkdn−kvdxn−k.

Now dkdxke2x=2ke2x

and since sin3x=ei3x−e−i3x2i ,

then

dn−kdxn−ksin3x=−i2dn−kdxn−k(e3ix−e−3ix)=−i2((3i)n−ke3ix−(−3i)n−ke−3ix).

So the final answer is

dndxne2xsin3x=∑k=0n(nk)2ke2x−i2((3i)n−ke3ix−(−3i)n−ke−3ix)

=−ie2x∑k=0n(nk)2k−1((3i)n−ke3ix−(−3i)n−ke−3ix